POJ 3695 Rectangles（容斥）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
#define MAXN 111111
#define inf 100000011
int 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;
}