2014台州学院ACM集训队寒假练习赛1

A Bridge

只想到一种情况另外一种情况是参考别人的

2个人为一轮

a b c d

a b 去 a 回 c d 去 b 回（最小的2个去 2个在回来。。。）

其实还有一种

a b 去 a 回 a c 去 a 回（只是最小的来回）

2中情况要具体分析不能全是第一种全是第二种每2个2个人取小值

这个真不会

语言: C++   用户:614173971  提交时间 : 2014-01-15 16:33:33.0#include <cstdio>
#include <algorithm>
using namespace std;
const __int64 MAX = 100010;
__int64 a[MAX];int main()
{__int64 n;__int64 i;while(~scanf("%I64d",&n)){for(i = 1;i <= n; i++)scanf("%I64d",&a[i]);if(n == 0){printf("%0\n");continue;}if(n == 1){printf("%I64d\n",a[1]);continue;}sort(a+1,a+n+1);__int64 sum = 0;for(i = n; i >= 4; i -= 2)sum += min(2 * a[1] + a[i] + a[i-1],a[1] + 2 * a[2] + a[i]);if(n%2 == 0)sum += a[2];else         sum += a[1] + a[2] + a[3];printf("%I64d\n",sum);}return 0;
}

B 3-sided dice

这题是看书学习的据说是经典题黑书上有讲转换成点是否在三角形内三维的叉积点积

好题

#include <cstring>
#include <cmath>
#include <cstdio>const double eps = 1e-10;
struct Point3
{double x, y, z;Point3(double x=0, double y=0, double z=0):x(x),y(y),z(z){}
};
typedef Point3 Vector3;Vector3 operator + (Vector3 A, Vector3 B)
{return Vector3(A.x+B.x, A.y+B.y, A.z+B.z);
}
Vector3 operator - (Vector3 A, Vector3 B)
{return Vector3(A.x-B.x, A.y-B.y, A.z-B.z);
}
int dcmp(double x)
{if(fabs(x) < eps)return 0;elsereturn x < 0 ? -1 : 1;
}
double Dot(Vector3 A, Vector3 B)//点积
{return A.x*B.x + A.y*B.y + A.z*B.z;
}
double Length(Vector3 A)//向量长度
{return sqrt(Dot(A, A));
}
Vector3 Cross(Vector3 A, Vector3 B)//求叉积
{return Vector3(A.y*B.z - A.z*B.y, A.z*B.x - A.x*B.z, A.x*B.y - A.y*B.x);
}
double Area(Point3 p0, Point3 p1, Point3 p2)
{return Length(Cross(p1-p0, p2-p0));
}
bool PointInTri(Point3 p, Point3 p0, Point3 p1, Point3 p2)//判断p是否在三角形内
{double area1 = Area(p, p0, p1);double area2 = Area(p, p1, p2);double area3 = Area(p, p2, p0);if(Area(p0,p1,p2) == 0)//3个点没有成为三角形 线段或者点的情况 {if(Dot(p0-p, p1-p) < 0)return true;if(Dot(p1-p, p2-p) < 0)return true;if(Dot(p2-p, p0-p) < 0)return true;return false;}if(dcmp(area1) == 0)//不能在三角形边上 return false;if(dcmp(area2) == 0)return false;if(dcmp(area3) == 0)return false;return dcmp(area1 + area2 + area3 - Area(p0,p1,p2)) == 0;
}bool ok(Point3 p, Point3 p0, Point3 p1, Point3 p2)//判断p是否在三角形所在的平面上
{Vector3 A = Cross(p1-p0, p2-p0);Vector3 B = p-p0;return dcmp(Dot(A, B)) == 0;
}
int main()
{double x,y,z;while(scanf("%lf %lf %lf",&x,&y,&z),x||y||z){Point3 p0(x,y,z);scanf("%lf %lf %lf",&x,&y,&z);Point3 p1(x,y,z);scanf("%lf %lf %lf",&x,&y,&z);Point3 p2(x,y,z);scanf("%lf %lf %lf",&x,&y,&z);Point3 p(x,y,z);if(ok(p,p0,p1,p2) && PointInTri(p,p0,p1,p2))puts("YES");elseputs("NO");}return 0;
}

C Relocation

dijkstra 因为k最多就5个求出k个超市到所有点的距离然后枚举起点（除去k个点）途径k个超市在回来你们所以需要的条件已经通过dijkstra求出来了

只需枚举求最小值 n*2^k 加上最短路 k*nlogn


#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
#define pii pair<int,int>
#define INF 999999999;
using namespace std;
const int MAX = 10010;vector <pii> edge[MAX];
int ok[MAX];
int dis[10][MAX];
int ans[10];
int n,m,k;void dijkstra(int s, int pos)
{priority_queue <pii, vector<pii> ,greater<pii> > q;bool vis[MAX];for(int i = 1;i <= n; i++){if(i == s)dis[pos][i] = 0;elsedis[pos][i] = INF;}memset(vis,false,sizeof(vis));q.push(make_pair(0,s));while(!q.empty()){pii u = q.top();q.pop();int x = u.second;if(vis[x])continue;vis[x] = true;int len = edge[x].size();for(int i = 0; i < len; i++){pii v = edge[x][i];if(dis[pos][v.second] > dis[pos][x] + v.first){dis[pos][v.second] = dis[pos][x] + v.first;q.push(make_pair(dis[pos][v.second],v.second));}}}
}int main()
{int i,j;int u,v,w;int sum = INF;scanf("%d %d %d",&n,&m,&k);for(i = 1;i <= k; i++){scanf("%d",&j);ans[i] = j;ok[j] = i;}for(i = 1;i <= m; i++){scanf("%d %d %d",&u,&v,&w);edge[u].push_back(make_pair(w,v));edge[v].push_back(make_pair(w,u));}for(i = 1; i <= k; i++)dijkstra(ans[i],i);for(i = 1;i <= n; i++){if(ok[i])continue;sort(ans+1,ans+1+k);do{int cnt = 0;for(j = 1;j <= k; j++){if(j == 1)cnt += dis[ok[ans[j]]][i];else if(j == k)cnt += dis[ok[ans[j]]][i] + dis[ok[ans[j]]][ans[j-1]];else cnt += dis[ok[ans[j]]][ans[j-1]];}if(sum > cnt)sum = cnt;//printf("%d\n",cnt);}while(next_permutation(ans+1,ans+k+1));}printf("%d\n",sum);return 0;
}

D Treasure Hunt

找出循环节就行了自己就是速度太慢了简单题也写了半天

#include <cstdio>
#include <cstring>
#include <cstdio>
using namespace std;
struct node
{int u;int v;
}a[220][22];
int b[100000];
int mp[220][22];
int main()
{int n,m,t,s;int i,j,u,v;int cnt;while(scanf("%d %d %d %d",&m,&t,&s,&n)!=EOF){cnt = 1;memset(mp,0,sizeof(mp));for(i = 1;i <= m; i++){for(j = 1;j <= t; j++){scanf("%d %d",&u,&v);a[i][j].u = v;a[i][j].v = u;}}u = s;v = 1;mp[s][1] = cnt;b[cnt++] = s;while(1){int uu = u;int vv = v;u = a[uu][vv].u;v = a[uu][vv].v;//printf("  %d %d\n",u,v);if(cnt-1 == n){printf("%d\n",b[cnt-1]);break;}if(mp[u][v]){int k = cnt - mp[u][v];int nn = (n - mp[u][v] + 1) % k;if(nn == 0)nn = k;printf("%d\n",b[mp[u][v] - 1 + nn]);//printf("%d\n",k);break;}else{mp[u][v] = cnt;b[cnt++] = u;}}          }return 0;
}

E The End of Corruption

我直接用了优先队列 1A 是不是数据水了啊 5s 用了340ms

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;int main()
{int n,m;int i,x;while(scanf("%d %d",&n,&m)!=EOF){priority_queue <int> q;for(i = 1;i <= n + m; i++){scanf("%d",&x);if(x == -1){printf("%d\n",q.top());q.pop();}elseq.push(x);}}return 0;
}

F What is the Rank?

一看是二分开始错了2次犯傻了

每次二分到那个位置把那个数插进去数组很难能实现 10秒我就乱搞了用vector insert 结果对了

#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAX = 45010;
int a[MAX];
vector <int> dp;
int pos[MAX];
int main()
{int n;int i,j,l,r,m,len;scanf("%d",&n);for(i = 0;i < n; i++)scanf("%d",&a[i]);dp.push_back(a[0]);pos[0] = 0;len = 0;for(i = 1;i < n; i++){l = 0;r = dp.size() - 1;while(l <= r){m = (l + r) >> 1;if(a[i] < dp[m])l = m + 1;elser = m - 1;}dp.insert(dp.begin()+l,a[i]);pos[i] = l;}for(i = 0;i < n; i++)printf("%d\n",pos[i]+1);return 0;
}

G Critical Intersections

一看题意发现是割点书上看到过学了下一下就懂了裸的应用可以看一下白皮书

一个dfs算法解决

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;const int MAX = 310;
vector <int> G[MAX];
int pre[MAX];
int low[MAX];
bool iscnt[MAX];
int n,m;
int cnt = 0;//时间戳
int dfs(int u,int fa)
{int lowu = pre[u] = ++cnt;int child = 0;//子节点数for(int i = 0; i < G[u].size(); i++){int v = G[u][i];if(!pre[v]){child++;int lowv = dfs(v,u);lowu = min(lowu,lowv);if(lowv >= pre[u])iscnt[u] = true;}else if(pre[v] < pre[u] && v != fa){lowu = min(lowu,pre[v]);}}if(fa < 0 && child == 1)iscnt[u] = false;low[u] = lowu;return lowu;
}
int main()
{int i,u,v;scanf("%d %d",&n,&m);for(i = 1;i <= m; i++){scanf("%d %d",&u,&v);G[u].push_back(v);G[v].push_back(u);}dfs(1,-1);int sum = 0;for(i = 1;i <= n; i++)if(iscnt[i])sum++;printf("%d\n",sum);return 0;
}

H Word Hopping

记忆化搜索就滑雪那样的没有了4个方向更简单了

题意看清就行

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char a[120][20];
int dp[120];
int vis[120];
int n;
bool ok1(char *s1,char *s2)
{int i,j;int cnt = 0;int len = strlen(s1);for(i = 0;i < len; i++){if(s1[i] > s2[i])return false;if(s1[i] < s2[i]){for(j = i + 1;j < len; j++)if(s2[i] == s1[j])break;if(j == len)return false;}if(s1[i] != s2[i])cnt++;}if(cnt != 1)return false;return true;
}bool ok2(char *s1,char *s2)
{int i,j;int cnt = 0;int len = strlen(s2);for(i = 0;i < len; i++){if(s1[i] != s2[i])break;}if(i == len)return true;for(j = i,i = j + 1; j < len; i++,j++)if(s1[i] != s2[j])return false;return true;
}
bool check(char *s1,char *s2)
{if(strlen(s1) < strlen(s2))return false;else if(strlen(s1) == strlen(s2)){if(ok1(s1,s2))return true;}else if(strlen(s1) - 1 == strlen(s2)){if(ok2(s1,s2))return true;}return false;
}int dfs(int u)
{if(vis[u])return dp[u];int i;int max = 0;for(i = 1;i <= n; i++){if(u == i)continue;if(!check(a[u],a[i]))continue;int temp = dfs(i);if(max < temp)max = temp;}dp[u] = max + 1;vis[u] = true;return dp[u];
}
int main()
{int i,j;while(scanf("%d",&n)!=EOF){for(i = 1;i <= n; i++)scanf("%s",a[i]);memset(dp,0,sizeof(dp));memset(vis,false,sizeof(vis));int max = 0;for(i = 1;i <= n; i++){int temp = dfs(i);if(max < temp)max = temp;}printf("%d\n",max);}return 0;
}

I Siruseri Metro System

不会啊求告知

J Equal Gifts

二维01背包一不小心A了一维也可以的

#include <cstdio>
#include <algorithm>
using namespace std;
struct node
{int u;int v;
}a[200];
bool dp[2][100000];
bool map[2][100000];
int main()
{int n,i,j,k;int m = 0;int sum = 0;while(scanf("%d",&n)!=EOF){memset(dp,false,sizeof(dp));memset(map,false,sizeof(map));sum = 0;for(i = 1;i <= n; i++){scanf("%d %d",&a[i].u,&a[i].v);sum += a[i].u + a[i].v;}map[1][0] = map[0][0] = true;for(i = 1;i <= n; i++){memcpy(dp,map,sizeof(map));memset(map,false,sizeof(map));for(j = m; j >= 0;j--){if(dp[0][j]){map[0][j+a[i].u] = true;map[0][j+a[i].v] = true;m = max(m,j+a[i].u);m = max(m,j+a[i].v);}if(dp[1][j]){map[1][j+a[i].u] = true;map[1][j+a[i].v] = true;m = max(m,j+a[i].u);m = max(m,j+a[i].v);}}}for(i = sum/2 ;i >= 1; i--){if(map[0][i] || map[1][i])break;}printf("%d\n",sum-2*i);}return 0;
}

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