hdu 2461 容斥原理 Rectangle

題意：http://acm.hdu.edu.cn/showproblem.php?pid=2461

        給你n（n很小）個長方形，求這中間任意長方形的面積並。

題解：

        容斥原理。

#include <cstdio>#include <iostream>#include <cstring>using namespace std;int n,Q;struct Rec{    int x1,x2,y1,y2;}rec[30];int s[(1 << 20) + 3], a[22],cnt,add,id;Rec intersec(Rec a, Rec b){    Rec ret;    if(a.x2 <= b.x1 || b.x2 <= b.x1 || a.y2 <= b.y1 || b.y2 <= a.y1){        ret.x1 = ret.y1 = ret.x2 = ret.y2 = 0;        return ret;    }    ret.x1 = max(a.x1, b.x1);    ret.y1 = max(a.y1, b.y1);    ret.x2 = min(a.x2, b.x2);    ret.y2 = min(a.y2, b.y2);    return ret;}int Area(Rec r){    if(r.x1 >= r.x2 || r.y1 >= r.y2)        return 0;    return (r.y2 - r.y1) * (r.x2 - r.x1);}int In_exclusion(int k, Rec r){    if(Area(r) == 0) return 0;//    int ret = 0;    for(int i = k; i <= cnt; i ++){        Rec tmp = intersec(rec[a[i]],r);        ret += Area(tmp) - In_exclusion(i + 1,tmp);    }    return ret;}int main(){    int k,ca = 1;    while(~scanf("%d%d",&n,&Q) && (n || Q)){        printf("Case %d:\n",ca ++);        Rec total ;        total.x1 = total.y1 = 0;        total.x2 = total.y2 = 1000;        memset(s,0,sizeof(s));        for(int i = 1; i <= n; ++i){            scanf("%d%d%d%d",&rec[i].x1,&rec[i].y1,&rec[i].x2,&rec[i].y2);        }        for(int qu = 1; qu <= Q; qu ++){            scanf("%d",&k);            cnt = 0, add = 0;            memset(a,0,sizeof(a));            while(k --){                scanf("%d",&id);                add = (add | (1 << (id - 1)));                a[cnt ++] = id;            }            if(s[add] == 0)                s[add] = In_exclusion(0,total);            printf("Query %d: %d\n",qu, s[add]);        }        printf("\n");    }    return 0;}