山东省第三届ACM省赛

ID	Title	Hint
A	Impasse (+)
B	Chess
C	An interesting game	最小费用最大流
D	n a^o7 !
E	Fruit Ninja I	dp
F	Pixel density	无
G	Mine Number	dfs
H	The Best Seat in ACM Contest	无
I	Pick apples	贪心，背包
J	Fruit Ninja II	积分

d.国际象棋？就是求能被0~16个棋子吃掉的棋子分别有几个。

s.当时有点思路，感觉对。但是时间好像不大够，也没敲。

用二分查找，查询周围16个位置有棋子没，能不能把它吃掉。

棋子个数n <= 10^5。

10^5*16*log(10^5)<=4*10^7

s.这个题感觉也可以，不知道为啥wa了。

感觉直接暴力贪心就行啊，题意理解错了吗？

c.上个刁丝测试代码。。。数据比较大的时候可以找到不同的，但是还是不懂为什么贪心不行。。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include<string.h>
#include <vector>
#include <queue>
#include<time.h>
using namespace std;
const int maxn=2200;
const int oo=0x3f3f3f3f;
struct Edge
{int u, v, cap, flow, cost;Edge(int u, int v, int cap, int flow, int cost):u(u), v(v), cap(cap), flow(flow), cost(cost) {}
};
struct MCMF
{int n, m, s, t;vector<Edge> edge;vector<int> G[maxn];int inq[maxn], d[maxn], p[maxn], a[maxn];void init(int n){this->n=n;for(int i=0; i<n; i++)G[i].clear();edge.clear();}void AddEdge(int u, int v, int cap, int cost){edge.push_back(Edge(u, v, cap, 0, cost));edge.push_back(Edge(v, u, 0, 0, -cost));m=edge.size();G[u].push_back(m-2);G[v].push_back(m-1);}bool spfa(int s, int t, int& flow, int& cost){memset(d, 0x3f, sizeof d);memset(inq, 0, sizeof inq);d[s]=0, inq[s]=1, p[s]=0, a[s]=oo;queue<int> q;q.push(s);while(!q.empty()){int u=q.front();q.pop();inq[u]=0;for(int i=0; i<G[u].size(); i++){Edge& e=edge[G[u][i]];if(e.cap>e.flow && d[e.v]>d[u]+e.cost){d[e.v]=d[u]+e.cost;p[e.v]=G[u][i];a[e.v]=min(a[u], e.cap-e.flow);if(!inq[e.v]){q.push(e.v);inq[e.v]=1;}}}}if(d[t]==oo)return false;flow+=a[t];cost+=d[t]*a[t];int u=t;while(u!=s){edge[p[u]].flow+=a[t];edge[p[u]^1].flow-=a[t];u=edge[p[u]].u;}return true;}int MinCost(int s, int t){int flow=0, cost=0;while(spfa(s, t, flow, cost));return cost;}
} net;int a[maxn], b[maxn];struct node{int h;int id;int h2;
}a2[1005],a3[1005];bool cmp(node a,node b){if(a.h!=b.h)return a.h<b.h;return a.h2<b.h2;
}
bool cmp2(node a,node b){if(a.h!=b.h)return a.h<b.h;return a.h2<b.h2;
}int test1[]={14,20,20,3,10,28,1,19,3,26,29,17,27,11,5,24,3,20,2,8,8,0,11,0,19,29,
10,23,17,3,14,28,14,8,6,6,8,5,15,1,23,28,17,9,15,20,13,6,3,11,8,6,23,3,1,21,13,20,17,0,16,
14,4,1,18,27,25,21,29,13,19,11,10,22,2,27,0,19,1,8,7,18,7,5,11,22,23,24,22,3,27,3,
17,21,26,8,17,17,17,19,12,28,14,17,28,19,25,12,10,1,30,6,1,1,29,16,0,6,22,27,25,
25,3,0,0,29,8,29,19,24,9,27,15,1,11,28,28,7,0,24,8,17,16,13,14,29,2,12,23,22,12,
25,13,19,21,11,9,10,9,16,16,13,20,22,10,30,29,27,12,23,16,11,22,24,9,15,4,26,4,15,
19,16,16,19,27,27,30,11,27,27,4,22,19,24,29,10,28,27,2,30,29,19,3,8,21,4,2,4,2,
23,13,6,20,21,5,10,2,8,3,2,3,17,10,7,25,16,13,12,10,23,30,27,22,24,6,10,0,0,30,
7,0,1,1,18,29,5,19,10,8,30,30,0,15,3,17,2,12,22,1,8,20,14,2,15,10,17,4,26,23,19,
15,25,30,17,23,13,18,24,15,1,15,14,5,13,18,10,28,11,11,30,15,0,2,28,4,27,29,21,30,
1,29,8,6,1,21,13,8,29,24,13,22,1,3,20,8,7,25,26,21,11,3,3,10,9,24,27,5,5,7,9,16,
17,18,0,5,9,12,10,29,9,20,24,11,4,3,30,12,30,26,6,17,14,20,15,26,17,27,28,15,21,
15,27,12,24,25,24,17,3,25,9,0,3,15,12,28,6,0,3,17,27,7,26,10,9,13,24,4,21,21,19,
20,29,15,15,0,24,22,17,30,2,6,21,12,21,16,24,4,10,25,6,18,26,19,5,7,17,18,7,1,
12,13,15,17,21,14,13,17,9,14,4,6,1,13,12,28,1,0,11,29,22,3,25,3,11,21,12,22,5,18,7,
5,0,12,26,11,17,8,14,0,18,6,7,8,30,23,19,0,26,6,15,23,24,30,29,11,20,2,21,14,
4,20,24,1,15,28,30,14,19,8,22,2,23,3,14,20,30,1,1,1,12,17,12,0,2,1,29,18,17,17,5,
17,21,8,8,23,15,11,2,18,30,12,26,9,13,15,11,30,12,23,21,5,18,12,4,20,8,15,19,19,
16,26,24,8,6,23,11,11,9,4,7,22,27,1,24,1,29,19,12,4,18,25,1,5,4,3,25,25,6,0,3,6,
23,23,15,4,6,21,10,20,25,19,11,4,26,4,28,8,8,22,26,12,6,20,28,24,7,4,11,25,28,23,
0,25,1,3,0,2,10,11,20,4,21,8,15,11,10,6,1,30,2,6,2,6,0,9,19,13,27,30,21,13,18,
17,10,21,27,23,2,5,13,15,10,4,4,18,10,9,5,7,10,24,19,12,2,12,26,6,25,8,18,3,13,19,
16,19,15,28,11,16,8,29,15,13,23,24,2,2,19,0,7,23,14,2,28,22,2,0,26,16,18,11,19,
0,10,19,5,26,8,8,8,13,5,30,20,29,21,30,3,7,29,27,19,28,13,2,15,18,28,13,5,24,13,
15,27,9,29,0,7,16,28,12,1,9,4,8,6,12,11,12,14,22,28,20,14,8,17,25,20,13,16,6,8,
3,7,7,28,17,23,17,13,16,0,19,23,5,17,6,20,21,9,14,30,2,26,7,5,30,24,2,10,23,22,8,
19,4,9,27,29,11,4,13,23,8,18,6,6,1,18,5,17,6,18,13,6,30,6,28,9,22,9,11,3,3,4,25,
29,3,10,19,26,14,11,29,9,23,4,0,23,29,14,2,21,29,18,10,10,3,2,4,16,20,23,2,5,0,
11,10,1,19,14,10,4,15,22,25,8,10,12,25,15,30,5,2,30,10,5,30,7,3,23,23,19,18,3,10,
13,2,30,16,30,2,6,22,30,3,4,28,6,2,22,3,3,5,20,14,16,12,14,28,12
};
int test2[]={0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,
3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,
4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6,
6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,
7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10,
10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,11,11,
11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,
12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,
13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,
14,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,
15,15,15,15,15,15,15,15,15,15,15,15,15,15,16,16,16,16,16,16,16,16,16,16,16,16,16,
16,16,16,16,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,17,17,17,17,17,17,
17,17,17,17,17,17,17,17,17,17,17,18,18,18,18,18,18,18,18,18,18,18,18,18,18,18,
18,18,18,18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,
19,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,
20,20,20,20,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21,21,21,21,21,21,21,21,
21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22,
22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23,
23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,
24,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25
,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,
27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,
28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29,
29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,30,
30,30,30,30,30,30,30,30,30,30,30,30,30
};int temp;int f(int N,int M,int K,int b[]){int T;//int N,M,K;//int a[1005],b[1005];int i;int ca=0;int sum;//int a2[1005],a3[1005];int p1,p2,p3,p4,q1,q2;int t1,t2;bool use[1005];//scanf("%d",&T);//while(T--){//scanf("%d%d%d",&N,&M,&K);//for(i=0;i<N;++i){//    scanf("%d",&a[i]);//}sum=0;for(i=0;i<N-1;++i){sum=sum+abs(a[i]-a[i+1]);temp=sum;if(a[i]>a[i+1]){a2[i].h=a[i];a2[i].h2=a[i+1];a3[i].h=a[i+1];a3[i].h2=a[i];a2[i].id=i;a3[i].id=i;}else{a2[i].h=a[i+1];a2[i].h2=a[i];a3[i].h=a[i];a3[i].h2=a[i+1];a2[i].id=i;a3[i].id=i;}/*cout<<"i:  "<<i<<"  a2[i].h "<<a2[i].h<<endl;cout<<"i:  "<<i<<"  a2[i].h2 "<<a2[i].h2<<endl;*/}
/*i=1;cout<<"##"<<endl;cout<<"i:  "<<i<<"  a2[i].h "<<a2[i].h<<endl;cout<<"i:  "<<i<<"  a2[i].h2 "<<a2[i].h2<<endl;*//*i=1;cout<<"##"<<endl;cout<<"i:  "<<i<<"  a2[i].h "<<a2[i].h<<endl;cout<<"i:  "<<i<<"  a2[i].h2 "<<a2[i].h2<<endl;*/sort(a2,a2+(N-1),cmp);sort(a3,a3+(N-1),cmp2);
/*i=1;cout<<"##"<<endl;cout<<"i:  "<<i<<"  a2[i].h "<<a2[i].h<<endl;cout<<"i:  "<<i<<"  a2[i].h2 "<<a2[i].h2<<endl;*//*i=1;cout<<"##"<<endl;cout<<"i:  "<<i<<"  a2[i].h "<<a2[i].h<<endl;cout<<"i:  "<<i<<"  a2[i].h2 "<<a2[i].h2<<endl;*///for(i=0;i<M;++i){//    scanf("%d",&b[i]);//}sort(b,b+M);p1=0;p2=N-2;p3=0;p4=N-2;q1=0;q2=M-1;memset(use,false,sizeof(use));for(i=0;i<K;++i){if(p1<0||p2<0||p3<0||p4<0||q1<0||q2<0){cout<<"越界"<<endl;}if(p1>N-2||p2>N-2||p3>N-2||p4>N-2||q1>M-1||q2>M-1){cout<<"越界2"<<endl;}while(use[a2[p1].id]==true){++p1;}//cout<<"p1 "<<p1<<endl;t1=b[q2]-a2[p1].h;//cout<<"t1  "<<t1<<endl;if(t1<0){t1=0;}while(use[a3[p4].id]==true){--p4;}//cout<<"p4  :"<<p4<<endl;t2=a3[p4].h-b[q1];//cout<<"t2 "<<t2<<endl;if(t2<0){t2=0;}if(t1>t2){sum=sum+2*t1;use[a2[p1].id]=true;////cout<<a2[p1].id<<"###   t1:  "<<t1<<endl;++p1;--q2;}else{sum=sum+2*t2;use[a3[p4].id]=true;////cout<<a3[p4].id<<"else   t2:   "<<t2<<endl;--p4;++q1;}//cout<<"i:"<<i<<",sum:"<<sum<<endl;
        }//printf("Case %d: %d\n",++ca,sum);return sum;// }

}int main()
{int ccc[maxn];int tot;int T, kase=0;//scanf("%d", &T);T=100;T=100;srand(time(NULL));while(T--){int n, m, k, ans=0;n=3;m=2;k=2;n=900;m=800;k=800;//scanf("%d%d%d", &n, &m, &k);net.init(n+100);memset(b, 0, sizeof b);for(int i=0; i<n; i++){a[i]=rand()%31;//a[i]=test1[i];//scanf("%d", a+i);
        }tot=0;for(int i=0; i<m; i++){int x;
/*x=rand()%31;//scanf("%d", &x);b[x]++;ccc[tot++]=x;*///          /*int fuck=rand()%31;b[fuck]++;ccc[tot++]=fuck;//        */
}for(int i=1; i<n; i++){ans+=abs(a[i]-a[i-1]);net.AddEdge(0, i, 1, 0);for(int j=0; j<=30; j++)if(b[j]){int dis=abs(a[i]-j)+abs(a[i-1]-j)-abs(a[i]-a[i-1]);net.AddEdge(i, n+j, 1, -dis);}}int S=n+50, T=S+1;for(int i=0; i<=30; i++)if(b[i])net.AddEdge(i+n, T, b[i], 0);net.AddEdge(S, 0, k, 0);int ans2=net.MinCost(S, T);int ans1=ans-ans2;//printf("Case %d: %d\n", ++kase, );//cout<<"贪心："<<endl;int tanxin=f(n,m,k,ccc);if(ans1!=tanxin){cout<<"ans:"<<ans<<endl;cout<<"ans2:"<<ans2<<endl;cout<<"temp:"<<temp<<endl;cout<<tanxin-temp<<endl;cout<<"yes"<<endl;cout<<"a:"<<endl;for(int i=0;i<n;++i){cout<<a[i]<<',';}cout<<endl;cout<<"ccc:"<<endl;for(int i=0;i<m;++i){cout<<ccc[i]<<',';}cout<<endl;cout<<endl;cout<<ans1<<"网络流"<<endl;cout<<tanxin<<"贪心"<<endl;//return 0;
}}return 0;
}
/*
yes
a:
14,20,20,3,10,28,1,19,3,26,29,17,27,11,5,24,3,20,2,8,8,0,11,0,19,29,10,23,17,3,1
4,28,14,8,6,6,8,5,15,1,23,28,17,9,15,20,13,6,3,11,8,6,23,3,1,21,13,20,17,0,16,14
,4,1,18,27,25,21,29,13,19,11,10,22,2,27,0,19,1,8,7,18,7,5,11,22,23,24,22,3,27,3,
17,21,26,8,17,17,17,19,12,28,14,17,28,19,25,12,10,1,30,6,1,1,29,16,0,6,22,27,25,
25,3,0,0,29,8,29,19,24,9,27,15,1,11,28,28,7,0,24,8,17,16,13,14,29,2,12,23,22,12,
25,13,19,21,11,9,10,9,16,16,13,20,22,10,30,29,27,12,23,16,11,22,24,9,15,4,26,4,1
5,19,16,16,19,27,27,30,11,27,27,4,22,19,24,29,10,28,27,2,30,29,19,3,8,21,4,2,4,2
,23,13,6,20,21,5,10,2,8,3,2,3,17,10,7,25,16,13,12,10,23,30,27,22,24,6,10,0,0,30,
7,0,1,1,18,29,5,19,10,8,30,30,0,15,3,17,2,12,22,1,8,20,14,2,15,10,17,4,26,23,19,
15,25,30,17,23,13,18,24,15,1,15,14,5,13,18,10,28,11,11,30,15,0,2,28,4,27,29,21,3
0,1,29,8,6,1,21,13,8,29,24,13,22,1,3,20,8,7,25,26,21,11,3,3,10,9,24,27,5,5,7,9,1
6,17,18,0,5,9,12,10,29,9,20,24,11,4,3,30,12,30,26,6,17,14,20,15,26,17,27,28,15,2
1,15,27,12,24,25,24,17,3,25,9,0,3,15,12,28,6,0,3,17,27,7,26,10,9,13,24,4,21,21,1
9,20,29,15,15,0,24,22,17,30,2,6,21,12,21,16,24,4,10,25,6,18,26,19,5,7,17,18,7,1,
12,13,15,17,21,14,13,17,9,14,4,6,1,13,12,28,1,0,11,29,22,3,25,3,11,21,12,22,5,18
,7,5,0,12,26,11,17,8,14,0,18,6,7,8,30,23,19,0,26,6,15,23,24,30,29,11,20,2,21,14,
4,20,24,1,15,28,30,14,19,8,22,2,23,3,14,20,30,1,1,1,12,17,12,0,2,1,29,18,17,17,5
,17,21,8,8,23,15,11,2,18,30,12,26,9,13,15,11,30,12,23,21,5,18,12,4,20,8,15,19,19
,16,26,24,8,6,23,11,11,9,4,7,22,27,1,24,1,29,19,12,4,18,25,1,5,4,3,25,25,6,0,3,6
,23,23,15,4,6,21,10,20,25,19,11,4,26,4,28,8,8,22,26,12,6,20,28,24,7,4,11,25,28,2
3,0,25,1,3,0,2,10,11,20,4,21,8,15,11,10,6,1,30,2,6,2,6,0,9,19,13,27,30,21,13,18,
17,10,21,27,23,2,5,13,15,10,4,4,18,10,9,5,7,10,24,19,12,2,12,26,6,25,8,18,3,13,1
9,16,19,15,28,11,16,8,29,15,13,23,24,2,2,19,0,7,23,14,2,28,22,2,0,26,16,18,11,19
,0,10,19,5,26,8,8,8,13,5,30,20,29,21,30,3,7,29,27,19,28,13,2,15,18,28,13,5,24,13
,15,27,9,29,0,7,16,28,12,1,9,4,8,6,12,11,12,14,22,28,20,14,8,17,25,20,13,16,6,8,
3,7,7,28,17,23,17,13,16,0,19,23,5,17,6,20,21,9,14,30,2,26,7,5,30,24,2,10,23,22,8
,19,4,9,27,29,11,4,13,23,8,18,6,6,1,18,5,17,6,18,13,6,30,6,28,9,22,9,11,3,3,4,25
,29,3,10,19,26,14,11,29,9,23,4,0,23,29,14,2,21,29,18,10,10,3,2,4,16,20,23,2,5,0,
11,10,1,19,14,10,4,15,22,25,8,10,12,25,15,30,5,2,30,10,5,30,7,3,23,23,19,18,3,10
,13,2,30,16,30,2,6,22,30,3,4,28,6,2,22,3,3,5,20,14,16,12,14,28,12,
ccc:
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,
3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,
4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,6,6,
6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,
7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10
,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,11,11,1
1,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,
12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13
,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,14,1
4,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,15,
15,15,15,15,15,15,15,15,15,15,15,15,15,15,16,16,16,16,16,16,16,16,16,16,16,16,16
,16,16,16,16,16,16,16,16,16,16,16,16,17,17,17,17,17,17,17,17,17,17,17,17,17,17,1
7,17,17,17,17,17,17,17,17,17,17,17,18,18,18,18,18,18,18,18,18,18,18,18,18,18,18,
18,18,18,18,18,18,18,18,18,18,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19,19
,19,19,19,19,19,19,19,19,19,19,19,20,20,20,20,20,20,20,20,20,20,20,20,20,20,20,2
0,20,20,20,20,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21,21,21,21,21,21,21,21,
21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,21,22,22,22,22,22,22
,22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,22,23,23,23,23,23,23,23,23,23,23,2
3,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,
24,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25
,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,26,2
7,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,27,28,28,
28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,29,29,29,29,29,29,29,29,29
,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,30,30,30,30,30,30,30,30,30,30,3
0,30,30,30,30,30,30,30,30,30,30,30,30,30,25090网络流
25086贪心Process returned 0 (0x0)   execution time : 1.981 s
Press any key to continue.*//*
#include <iostream>
#include <cmath>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <list>
#include <algorithm>
#include <map>
#include <set>
#define LL long long
#define Pr pair<int,int>
#define fread() freopen("data1.in","r",stdin)
#define fwrite() freopen("out.out","w",stdout)using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;struct Edge
{int v,f,w,next;
};Edge eg[66666];
int head[2333],h[2333],hcnt[2333];
int dis[2333],pre[2333];
bool vis[2333];
int tp,ans,minf;void Add(int u,int v,int w,int f)
{eg[tp].v = v;eg[tp].w = w;eg[tp].f = f;eg[tp].next = head[u];head[u] = tp++;
}bool bfs(int st,int en)
{memset(dis,-1,sizeof(dis));memset(vis,0,sizeof(vis));memset(pre,-1,sizeof(pre));queue <int> q;q.push(st);dis[st] = 0;minf = INF;int u,v,w,f;while(!q.empty()){u = q.front();q.pop();vis[u] = 0;for(int i = head[u]; i != -1; i = eg[i].next){v = eg[i].v;w = eg[i].w;f = eg[i].f;if(f && (dis[v] == -1 || dis[v] < dis[u]+w)){dis[v] = dis[u]+w;minf = min(f,minf);pre[v] = i;if(!vis[v]){q.push(v);vis[v] = 1;}}}}if(pre[en] == -1) return false;ans += dis[en];for(int i = pre[en]; i != -1; i = pre[eg[i^1].v]){eg[i].f -= minf;eg[i^1].f += minf;}return true;
}int main()
{int t,n,m,k,x;scanf("%d",&t);for(int z = 1; z <= t; ++z){scanf("%d%d%d",&n,&m,&k);memset(head,-1,sizeof(head));memset(hcnt,0,sizeof(hcnt));tp = 0;ans = 0;for(int i = 0; i < n; ++i){scanf("%d",&h[i]);if(i) ans += abs(h[i]-h[i-1]);}while(m--){scanf("%d",&x);hcnt[x]++;}//n为起点 0~n-2表示初始第i个山爬到i+1山路程 n+1~n+30表示高1~30的山数 n+31表示起点前的一个点(限流) n+32表示终点Add(n+33,n+31,0,k);Add(n+31,n+33,0,0);for(int i = 0; i <= 30; ++i){if(!hcnt[i]) continue;Add(n+31,n+i,0,hcnt[i]);Add(n+i,n+31,0,0);for(int j = 0; j < n-1; ++j){Add(n+i,j,abs(h[j]-i)+abs(h[j+1]-i)-abs(h[j]-h[j+1]),hcnt[i]);Add(j,n+i,-abs(h[j]-i)-abs(h[j+1]-i)+abs(h[j]-h[j+1]),0);}}for(int j = 0; j < n-1; ++j){Add(j,n+32,0,1);Add(n+32,j,0,0);}while(bfs(n+33,n+32));printf("Case %d: %d\n",z,ans);}return 0;
}*/

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s.正宗解法，最小费用流

参考：山东省第三届省赛 C 题 – An interesting game(费用流)

ps：参考的这个代码比较快，600ms。我的好像得1000多。。。

大意：有n个山坡，排成一排，为了增加难度，现在要从m个山坡中挑选k个，插入到原有的n个山坡中，两个原有的山坡中只能插入一个山坡，原有山坡的两侧不能插入。使插入k个山坡后的难度最大，总的难度是相邻两个山坡差值绝对值的和。
思路：训练赛的时候根本没往图论上想，当时想贪心不太可能，还以为是dp。
如果把原问题看做是原有山坡的排列中，两个相邻山坡中间的间隙和m个山坡做匹配的话，就可以看做是，最大流量为k时，最大费用是多少，费用流可以搞。对于最大费用，可以先把费用取负值，求出最小费用后再取负值，就是最大费用。匹配的费用是插入之后的难度的增加量，最大费用加上原山坡的难度，就是答案。
因为n和m的范围是[2, 1000]，直接建图会TLE，发现插入的m个山坡的范围只有[0, 30]，这么小的数据范围一定能优化答案，把插入山坡的高度看做结点，建图。
设置源点S，超级源点SS，汇点T。
S对所有间隙连边，容量1，费用0.
所有间隙对所有高度连边，容量1，费用是难度增加量的相反数.
所有高度对T连边，容量为此高度的个数，费用0.
SS对S连边，容量k，费用0.
答案就是原图的难度-最小费用。

c.最小费用最大流

/*
SPFA版费用流
最小费用最大流，求最大费用最大流只需要取相反数，结果取相反数即可。
点的总数为N，点的编号0~N-1
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;const int MAXN=1500;
const int MAXM=80000;
const int INF=0x3f3f3f3f;
struct Edge{int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从0~N-1
void init(int n){N=n;tol=0;memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost){edge[tol].to=v;edge[tol].cap=cap;edge[tol].cost=cost;edge[tol].flow=0;edge[tol].next=head[u];head[u]=tol++;edge[tol].to=u;edge[tol].cap=0;edge[tol].cost=-cost;edge[tol].flow=0;edge[tol].next=head[v];head[v]=tol++;
}
bool spfa(int s,int t){queue<int>q;for(int i=0;i<N;i++){dis[i]=INF;vis[i]=false;pre[i]=-1;}dis[s]=0;vis[s]=true;q.push(s);while(!q.empty()){int u=q.front();q.pop();vis[u]=false;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost){dis[v]=dis[u]+edge[i].cost;pre[v]=i;if(!vis[v]){vis[v]=true;q.push(v);}}}}if(pre[t]==-1)return false;else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s,int t,int &cost){int flow=0;cost=0;while(spfa(s,t)){int Min=INF;for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){if(Min>edge[i].cap-edge[i].flow)Min=edge[i].cap-edge[i].flow;}for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){edge[i].flow+=Min;edge[i^1].flow-=Min;cost+=edge[i].cost*Min;}flow+=Min;}return flow;
}int main(){int T;int N2,M,K;int X[1005],Y[1005];int YNum[40];int i,j;int sp,sp2;//源点，源点2int sc;//汇点int sum;int tmp;int mi_cost;int ma_flow;int ca=0;scanf("%d",&T);while(T--){scanf("%d%d%d",&N2,&M,&K);init(N2+50);for(i=0;i<N2;++i){scanf("%d",&X[i]);}memset(YNum,0,sizeof(YNum));for(i=0;i<M;++i){scanf("%d",&Y[i]);++YNum[Y[i]];}sum=0;for(i=0;i<N2-1;++i){//初始值sum=sum+abs(X[i]-X[i+1]);}sp=N2+40;for(i=0;i<N2-1;++i){//加边addedge(sp,i,1,0);//源点到空隙for(j=0;j<=30;++j){if(YNum[j]){tmp=abs(X[i]-j)+abs(X[i+1]-j)-abs(X[i]-X[i+1]);addedge(i,N2+j,1,-tmp);//空隙到M个山丘
                }}}sp2=N2+41;sc=N2+42;addedge(sp2,sp,K,0);for(i=0;i<=30;++i){if(YNum[i]){addedge(N2+i,sc,YNum[i],0);}}ma_flow=minCostMaxflow(sp2,sc,mi_cost);printf("Case %d: %d\n",++ca,sum-mi_cost);}return 0;
}

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d.好像是按要求输出串

s.没写这个题

c.张

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#define MAX 500
using namespace std;
int main ()
{char temp1[MAX],temp2[MAX],str[MAX];strcpy(temp1," n5!wpuea^o7!usimdnaevoli");strcpy(temp2," usimdnaevolin5!wpuea^o7!");int Len = strlen(temp1);int T;scanf("%d",&T);getchar();int coun=1;while(T--){gets(str);int len=strlen(str);for(int i=0,j=0; i<len; i++){for(j=0; j<Len; j++)if(str[i]==temp1[j])break;if(j<Len) str[i]=temp2[j];}printf("Case %d: ",coun++);for(int i=len-1; i>=0; i--)printf("%c",str[i]);printf("\n");}return 0;
}

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d.水果忍者，水果从上向下掉落，并且只能水平切。

给你n个水果的出现时间，出现的水平位置，和他是否是好水果，切到好水果+1，切到坏水果-1，连续切到三个以上好水果分数加倍。

两刀之间的间隔大于等于m，求能求得的最大分数。

s.先求出每一秒出手能得到的最大的得分，很简单。

再用DP求最大分数。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;#define MAXN 10010int d[MAXN][210];//d[i][j]标志i秒j位置水果
int dp[MAXN];//dp[i]表示前i秒获得的最大分数int main(){int T;int n,m;int t,s,p;int i,j;int ma_t;//最大时间int ma_p[MAXN];//每秒最大位置int sum,num,ma;//
    int ma_sum[MAXN];//每秒可获得的最大分数int ca=0;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);memset(d,-1,sizeof(d));ma_t=0;memset(ma_p,0,sizeof(ma_p));for(i=0;i<n;++i){scanf("%d%d%d",&t,&s,&p);d[t][p]=s;if(t>ma_t){ma_t=t;}if(p>ma_p[t]){ma_p[t]=p;}}memset(ma_sum,0,sizeof(ma_sum));for(i=1;i<=ma_t;++i){//先求出每秒可获得的最大分数sum=0;num=0;ma=0;for(j=1;j<=ma_p[i];++j){if(d[i][j]==0){++num;}else if(d[i][j]==1){if(num>=3){sum=sum+num*2;}else{sum=sum+num;}if(sum>ma){ma=sum;}num=0;--sum;if(sum<0){sum=0;}}}if(num>0){if(num>=3){sum=sum+num*2;}else{sum=sum+num;}if(sum>ma){ma=sum;}}ma_sum[i]=ma;}memset(dp,0,sizeof(dp));for(i=1;i<=m;++i){dp[i]=max(dp[i-1],ma_sum[i]);//不出手，和出手
        }for(i=m+1;i<=ma_t;++i){dp[i]=max(dp[i-1],dp[i-m-1]+ma_sum[i]);}printf("Case %d: %d\n",++ca,dp[ma_t]);}return 0;
}

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d.串中提取数字

s.比较坑的是里面空格只要一个就行了

c.当时臃肿的代码。虽然长，但是思路还可以

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;char str[12345];char str1[12345];
char str2[12345];char num1[12345];
char num2[12345];
char num3[12345];int main(){int T;int i;int len;int len1,len2;int p1,p2;int p3,p4;int len_num1,len_num2,len_num3;int j;double a,a1,a2,b,b1,b2,c,c1,c2;int p5;double Dp;double ans;int ca=0;int p6,p7;/*num1         num2 num3The new iPad  0009.7 inches 2048*1536 PADp3        p1         p2    p40009.7        2048*1536p5p6,07是考虑了后面的小数点，实际好像后面只能是整数*/scanf("%d",&T);getchar();while(T--){scanf("%[^\n]",str);getchar();len=strlen(str);len1=0;len2=0;len_num1=0;len_num2=0;len_num3=0;for(i=0;i<len;++i){if(str[i]=='i'){if(str[i+1]=='n'&&str[i+2]=='c'&&str[i+3]=='h'&&str[i+4]=='e'&&str[i+5]=='s'&&str[i+6]==' '){p1=i;j=i-1;while(str[j]==' '){--j;}while(str[j]!=' '){num1[len_num1++]=str[j--];}while(str[j]==' '){--j;}p3=j;}}if(str[i]=='*'){if(i>=1&&'0'<=str[i-1]&&str[i-1]<='9'){if('0'<=str[i+1]&&str[i+1]<='9'){p2=i;j=i-1;while(str[j]!=' '){num2[len_num2++]=str[j--];}j=i+1;while(str[j]!=' '){num3[len_num3++]=str[j++];}while(str[j]==' '){++j;}p4=j;}}}}for(i=0;i<=p3;++i){if(str[i]!=' '){break;}}str1[len1++]=str[i++];for(;i<=p3;++i){if(str[i]==' '&&str1[len1-1]==' '){continue;}str1[len1++]=str[i];}str1[len1]='\0';for(i=p4;i<len;++i){if(str[i]==' '&&str2[len2-1]==' '){continue;}if('a'<=str[i]&&str[i]<='z'){str2[len2++]=str[i];}else if('A'<=str[i]&&str[i]<='Z'){str2[len2++]=str[i]+32;}else{str2[len2++]=str[i];}}str2[len2]='\0';for(i=len2-1;i>=0;--i){if(str2[i]!=' '){len2=i+1;break;}}str2[i+1]='\0';p5=-1;for(i=0;i<len_num1;++i){if(num1[i]=='.'){p5=i;break;}}a=0;a1=0;a2=0;if(p5==-1){for(i=0;i<len_num1;++i){a1=a1+(num1[i]-'0')*pow(10,i);}}else{for(i=0;i<p5;++i){a2=a2/10+(num1[i]-'0')/10.0;}for(i=p5+1;i<len_num1;++i){a1=a1+(num1[i]-'0')*pow(10,(i-(p5+1)));}}a=a1+a2;if(a==0){printf("Case %d: The %s of %s's PPI is %.2f.\n",++ca,str2,str1,0.0);continue;}p6=-1;for(i=0;i<len_num2;++i){if(num2[i]=='.'){p6=i;break;}}if(p6==-1){b=0;for(i=0;i<len_num2;++i){b=b+(num2[i]-'0')*pow(10,i);}}else{b=0;b1=0;b2=0;for(i=0;i<p6;++i){b2=b2/10+(num2[i]-'0')/10.0;}for(i=p6+1;i<len_num2;++i){b1=b1+(num2[i]-'0')*pow(10,(i-(p6+1)));}b=b1+b2;}p7=-1;for(i=0;i<len_num3;++i){if(num3[i]=='.'){p7=i;break;}}if(p7==-1){c=0;for(i=0;i<len_num3;++i){c=c*10+(num3[i]-'0');}}else{c=0;c1=0;c2=0;for(i=0;i<p7;++i){c1=c1*10+(num3[i]-'0');}for(i=p7+1;i<len_num3;++i){c2=c2/10+(num3[i]-'0')/10.0;}c=c1+c2;}Dp=sqrt(b*b+c*c);ans=Dp/a;printf("Case %d: The %s of %s's PPI is %.2f.\n",++ca,str2,str1,ans);}return 0;
}

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d.类似于扫雷的游戏，这里不是周围8个位置，而是上下左右加它自己5个位置。给出周围多少个雷的数组，求出雷的分布。

s.dfs。先确定第一行，下面的每个位置只看上面的位置即可。

#include<iostream>
#include<stdio.h>
using namespace std;int d[25][25];
char g[25][25];int n,m;
int dir[][2]={{-1,0},{1,0},{0,-1},{0,1},{0,0}};
bool stop;void dfs(int r,int c);void f1(int r,int c){int i;int a,b;g[r][c]='*';for(i=0;i<5;++i){a=r+dir[i][0];b=c+dir[i][1];if( 0<=a&&a<n && 0<=b&&b<m ){--d[a][b];}}if(c==m-1){a=r+1;b=0;}else{a=r;b=c+1;}dfs(a,b);g[r][c]='\0';for(i=0;i<5;++i){a=r+dir[i][0];b=c+dir[i][1];if( 0<=a&&a<n && 0<=b&&b<m ){++d[a][b];}}
}
void f2(int r,int c){int a,b;g[r][c]='.';if(c==m-1){a=r+1;b=0;}else{a=r;b=c+1;}dfs(a,b);g[r][c]='\0';
}void dfs(int r,int c){int i,j;if(r==n){for(i=0;i<m;++i){if(d[n-1][i]!=0){return;}}stop=true;for(i=0;i<n;++i){for(j=0;j<m;++j){printf("%c",g[i][j]);}printf("\n");}return;}int a,b;if(r==0){bool flag=true;//可以放雷for(i=0;i<5;++i){a=r+dir[i][0];b=c+dir[i][1];if( 0<=a&&a<n && 0<=b&&b<m ){if(d[a][b]==0){flag=false;break;}}}if(flag){f1(r,c);if(stop){return;}}f2(r,c);}else{if(d[r-1][c]==1){f1(r,c);if(stop){return;}}else if(d[r-1][c]==0){f2(r,c);}}
}int main(){int T;int i,j;int ca=0;char str[25];scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(i=0;i<n;++i){scanf("%s",str);for(j=0;j<m;++j){d[i][j]=str[j]-'0';}}printf("Case %d:\n",++ca);stop=false;dfs(0,0);}return 0;
}

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d.在一个N*M（N,M<=20）的矩阵中，某个位置的值取决于上下左右和它自己，求出所有位置的值。

s.双重循环，直接求一遍就行了。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};int main(){int T;int N,M;int s[25][25];int v[25][25];int i,j;int a,b;int ma,ma_i,ma_j;int ca=0;int k;scanf("%d",&T);while(T--){scanf("%d%d",&N,&M);for(i=0;i<N;++i){for(j=0;j<M;++j){scanf("%d",&s[i][j]);}}memset(v,0,sizeof(v));for(i=0;i<N;++i){for(j=0;j<M;++j){for(k=0;k<4;++k){a=i+dir[k][0];b=j+dir[k][1];if(a<0||b<0||a>=N||b>=M){v[i][j]-=1;continue;}if(s[a][b]>s[i][j]){v[i][j]=v[i][j]+(s[a][b]-s[i][j]);}else{v[i][j]=v[i][j]-(s[i][j]-s[a][b]);}}}}ma=v[0][0];ma_i=0;ma_j=0;for(i=0;i<N;++i){for(j=0;j<M;++j){if(v[i][j]>=ma){ma=v[i][j];ma_i=i;ma_j=j;}}}printf("Case %d: %d %d %d\n",++ca,ma,ma_i+1,ma_j+1);}return 0;
}

View Code

d.比较直接的一个完全背包，但是大小有这么大V <= 100,000,000

s.很显然，直接来会爆内存了。

但是这个题物品大小不大1 <= S<= 100。然后很萎缩的5000以上贪心，剩余的用完全背包。。。

为什么过了？

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;struct node{long long S;long long P;double q;
}a[3];bool cmp(node a,node b){return a.q>b.q;
}int main(){int T;long long V;long long sum;long long dp[6000];long long tmp;int i,j;int ca=0;scanf("%d",&T);while(T--){scanf("%lld%lld",&a[0].S,&a[0].P);a[0].q=(double)(a[0].P)/a[0].S;//cout<<a[0].q<<endl;scanf("%lld%lld",&a[1].S,&a[1].P);a[1].q=(double)(a[1].P)/a[1].S;//cout<<a[1].q<<endl;scanf("%lld%lld",&a[2].S,&a[2].P);a[2].q=(double)(a[2].P)/a[2].S;//cout<<a[2].q<<endl;scanf("%lld",&V);sort(a,a+3,cmp);sum=V%a[0].S;for(i=1;;++i){if(a[0].S*i>=5000){break;}}sum=sum+a[0].S*i;memset(dp,0,sizeof(dp));for(i=0;i<3;++i){for(j=a[i].S;j<=sum;++j){tmp=dp[j-a[i].S]+a[i].P;if(tmp>dp[j]){dp[j]=tmp;}}}for(i=sum;;--i){if(dp[i]>0){break;}}printf("Case %d: %lld\n",++ca,((V-sum)/a[0].S)*a[0].P+dp[i]);}return 0;
}

View Code

d.好像是积分求椭球体的体积。

c.王

#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;int main()
{int T;scanf("%d",&T);int i;for(i=1; i<=T; i++){int a,b,h;scanf("%d%d%d",&a,&b,&h);double V;V=(4.0/3)*M_PI*a*b*b;if(h>=b){printf("Case %d: %.3lf\n",i,V);}else{double v=M_PI*a*b*(b-h)-M_PI*(1.0*a/b)*(1.0/3)*(b*b*b-h*h*h);if(V-v-v>0)printf("Case %d: %.3lf\n",i,V-v);elseprintf("Case %d: %.3lf\n",i,v);}}return 0;
}

View Code

转载于:https://www.cnblogs.com/bofengyu/p/5327856.html

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