[CF4D]Mysterious Present

4D:Mysterious Present

题意简述

给出一个限制 >(w,h) >(w,h)和 n n个物品的二维信息(xi,yi)(x_i,y_i)。
求物品二维都满足限制的前提下，最长严格上升子序列 xi>xi−1,yi>yi−1 x_i>x_{i-1},y_i>y_{i-1}。

数据范围

1≤n≤ 5000 1 \leq n \leq 5000
1≤w,h≤106 1 \leq w,h \leq 10^6
1≤xi,yi≤106 1 \leq x_i,y_i \leq 10^6

思路

按照一维排序，第二维就是一个LIS。
注意二分的时候是lower_bound等于的也要包括。
时间复杂度 O(nlogn) O(nlogn)

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct data{int x,y;int id;bool operator < (const data &n1) const{return x==n1.x ? y>n1.y :x<n1.x;}
}d[5010];
int n,w,h,t,tail;
int g[5010],pre[5010];
int in()
{char ch=getchar();while (ch<'0'||ch>'9')ch=getchar();int ret=0;while (ch>='0'&&ch<='9')ret=ret*10+ch-'0',ch=getchar();return ret;
}
bool cmp(const int &a,const int &b)
{return d[a].y<d[b].y;
}
int dfs(int x)
{if (pre[x])dfs(pre[x]);printf("%d ",x);
}
int main()
{n=in(),w=in(),h=in();for (int i=1;i<=n;i++)d[i].x=in(),d[i].y=in(),d[i].id=i;sort(d+1,d+n+1);g[tail++]=0;for (int i=1;i<=n;i++){if (d[i].x<=w||d[i].y<=h)continue;if (d[i].y>d[g[tail-1]].y){g[tail++]=i;pre[d[i].id]=d[g[tail-2]].id;}else{t=lower_bound(g,g+tail,i,cmp)-g;g[t]=i;pre[d[i].id]=d[g[t-1]].id;}}printf("%d\n",tail-1);if (tail>1)dfs(d[g[tail-1]].id);return 0;
}

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[CF4D]Mysterious Present

4D:Mysterious Present

题意简述

数据范围

思路

代码

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