poj 3695 Rectangles(容斥原理)

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在容斥原理题单里看到这个题，第一想法肯定是扫描线啊。但一看题单分析，还真是容斥。矩形相交的图形和文氏图差不多。然后dfs容斥就好了
题单里第四题：http://blog.csdn.net/shengtao96/article/details/52490020

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;struct Rec{    int x1,y1,x2,y2;};Rec rec[30];int nums[30];int calc(Rec& t){    return (t.x2-t.x1)*(t.y2-t.y1);}bool Union(Rec& a, Rec b){    Rec t;    //左边界    t.x1 = max(a.x1,b.x1);    //下边界    t.y1 = max(a.y1,b.y1);    //右边界    t.x2 = min(a.x2,b.x2);    //上边界    t.y2 = min(a.y2,b.y2);    if(t.x1 < t.x2 && t.y1 < t.y2)    {        a = t;        return false;    }    return true;}int dfs(int n, Rec t, int num){    if(n == -1)    {        if(num)        {            if(num&1) return calc(t);            else return -1*calc(t);        }        return 0;    }    int ans = dfs(n-1,t,num);    //俩矩阵没交集就返回，有交集就继续    if(Union(t,rec[nums[n]-1])) return ans;    ans += dfs(n-1,t,num+1);    return ans;}int main(){    int n,m;    int time = 0;    while(scanf("%d %d",&n,&m) && n+m)    {        for(int i = 0; i < n; ++i)            scanf("%d %d %d %d",&rec[i].x1,&rec[i].y1,&rec[i].x2,&rec[i].y2);        int r,res;        printf("Case %d:\n",++time);        for(int i = 1; i <= m; ++i)        {            scanf("%d",&r);            for(int i = 0; i < r; ++i)                scanf("%d",&nums[i]);            Rec t;            t.x1 = t.y1 = -1;            t.x2 = t.y2 = 1001;            res = dfs(r-1,t,0);            printf("Query %d: %d\n",i,res);        }        printf("\n");    }    return 0;}

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