标签：scan find note ali you row second 交点 main
You are given nn rectangles on a plane with coordinates of their bottom left and upper right points. Some (n?1)(n?1) of the given nn rectangles have some common point. A point belongs to a rectangle if this point is strictly inside the rectangle or belongs to its boundary.
Find any point with integer coordinates that belongs to at least (n?1)(n?1) given rectangles.
The first line contains a single integer nn (2≤n≤1326742≤n≤132674) — the number of given rectangles.
Each the next nn lines contains four integers x1x1, y1y1, x2x2 and y2y2 (?109≤x1<x2≤109?109≤x1<x2≤109, ?109≤y1<y2≤109?109≤y1<y2≤109) — the coordinates of the bottom left and upper right corners of a rectangle.
Print two integers xx and yy — the coordinates of any point that belongs to at least (n?1)(n?1) given rectangles.
3
0 0 1 1
1 1 2 2
3 0 4 1
1 1
3
0 0 1 1
0 1 1 2
1 0 2 1
1 1
4
0 0 5 5
0 0 4 4
1 1 4 4
1 1 4 4
1 1
5
0 0 10 8
1 2 6 7
2 3 5 6
3 4 4 5
8 1 9 2
3 4
The picture below shows the rectangles in the first and second samples. The possible answers are highlighted.
题意：给出n个矩形，找一个点至少同时在n-1个矩形内。
思路：我们分别对每条对角线求前缀交和后缀交，则若在每个条对角线左右两边的的前缀与后缀取交后还存在交点，即为解。
代码：
1 #include"bits/stdc++.h" 2 3 #define db double 4 #define ll long long 5 #define vl vector<ll> 6 #define ci(x) scanf("%d",&x) 7 #define cd(x) scanf("%lf",&x) 8 #define cl(x) scanf("%lld",&x) 9 #define pi(x) printf("%d\n",x) 10 #define pd(x) printf("%f\n",x) 11 #define pl(x) printf("%lld\n",x) 12 #define rep(i, n) for(int i=0;i<n;i++) 13 using namespace std; 14 const int N = 1e6 + 5; 15 const int mod = 1e9 + 7; 16 const int MOD = 998244353; 17 const db PI = acos(-1.0); 18 const db eps = 1e-10; 19 const ll INF = 0x3fffffffffffffff; 20 int n; 21 struct P{ 22 int d,l,u,r; 23 inline P operator | (P a){ 24 return (P){max(a.d,d),max(a.l,l),min(a.u,u),min(a.r,r)}; 25 } 26 }a[N],pre[N],suf[N]; 27 int main(){ 28 ci(n); 29 for(int i=1;i<=n;i++){ 30 ci(a[i].d),ci(a[i].l),ci(a[i].u),ci(a[i].r); 31 } 32 pre[0]=suf[n+1]={-mod,-mod,mod,mod};//初始化 33 for(int i=1;i<=n;i++){ 34 pre[i]=pre[i-1]|a[i];//前缀 35 } 36 for(int i=n;i>=1;i--){ 37 suf[i]=suf[i+1]|a[i];//后缀 38 } 39 for(int i=1;i<=n;i++){ 40 P tmp=pre[i-1]|suf[i+1];//取交 41 if(tmp.d<=tmp.u&&tmp.l<=tmp.r) return !printf("%d %d\n",tmp.d,tmp.l); 42 } 43 return 0; 44 }
AIM Tech Round 5C. Rectangles 思维
标签：scan find note ali you row second 交点 main
原文地址：https://www.cnblogs.com/mj-liylho/p/9547575.html
