POJ 3695 Rectangles（容斥）

来源：互联网发布：linux nvidia 驱动编辑：程序博客网时间：2024/06/05 21:49

题意：

n个矩阵m次访问，求每次访问的总面积

参考http://http://blog.csdn.net/sdj222555/article/details/7983902

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>using namespace std;#define MAXN 111111#define inf 100000011int ask[MAXN],st[1111111];int n,m;struct Rec{    int x1,x2,y1,y2;}p[22];void dfs(int xa,int ya, int xb, int yb, int deep, int flag ,int sta){if(xa>=xb || ya>=yb) return;if(deep==n){if(sta){for(int i=1;i<=m;i++){if((ask[i]|sta)==ask[i])st[ask[i]] += flag*(xb-xa)*(yb-ya);}}return;}dfs(xa,ya,xb,yb,deep+1,flag,sta);dfs(max(xa,p[deep+1].x1),max(ya,p[deep+1].y1),min(xb,p[deep+1].x2),min(yb,p[deep+1].y2),deep+1,-flag,sta|(1<<deep));}int in(){    char ch;    int a = 0;    while((ch = getchar()) == ' ' || ch == '\n');    a += ch - '0';    while((ch = getchar()) != ' ' && ch != '\n')    {        a *= 10;        a += ch - '0';    }    return a;}int main(){int cas=1;while(cin>>n>>m){if(n==0 && m==0) break;memset(st,0,sizeof(st));memset(ask,0,sizeof(ask));for(int i=1;i<=n;i++){scanf("%d%d%d%d", &p[i].x1, &p[i].y1, &p[i].x2, &p[i].y2);}int t,x;for(int i=1;i<=m;i++){t = in();while(t--){x = in();ask[i] |= (1<<(x-1));}}dfs(0,0,inf,inf,0,-1,0);printf("Case %d:\n",cas++);for(int i=1;i<=m;i++){printf("Query %d: %d\n",i,st[ask[i]]);}printf("\n");}return 0;}

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