HDU 2461 Rectangles 容斥原理

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题意：坐标系上有N个矩形，给出了他们的左下顶点和右上顶点。每次操作是对其中的R个矩形进行染色，然后要你输出这些被染色的矩形的面积之和（重叠部分不能重复计算）。

题解：标准的容斥原理。

#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>using namespace std;struct Rectangle { int x1, y1, x2, y2;};Rectangle rec[30];int a[30], cnt;int s[1100000];int n, m;Rectangle Intersection ( Rectangle r1, Rectangle r2 ){    Rectangle ret;    if ( r1.x2 <= r2.x1 || r2.x2 <= r1.x1 || r1.y2 <= r2.y1 || r2.y2 <= r1.y1 )    {        ret.x1 = ret.y1 = ret.y1 = ret.y2 = 0;        return ret;    }    ret.x1 = max ( r1.x1, r2.x1 );    ret.y1 = max ( r1.y1, r2.y1 );    ret.x2 = min ( r1.x2, r2.x2 );    ret.y2 = min ( r1.y2, r2.y2 );    return ret;}int Area ( Rectangle 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, Rectangle r ){    if ( Area(r) == 0 ) return 0;    int ret = 0;    Rectangle tmp;    for ( int i = k; i < cnt; i++ )    {        tmp = Intersection(r,rec[a[i]]);        ret += Area(tmp) - In_Exclusion ( i + 1, tmp );    }    return ret;}int main(){    int t = 0;    Rectangle total;    total.x1 = total.y1 = 0;    total.x2 = total.y2 = 1000;    while (scanf("%d%d",&n,&m) && (m||n))    {        int i, R, id, add;        for ( i = 1; i <= n; i++ )            scanf("%d%d%d%d",&rec[i].x1,&rec[i].y1,&rec[i].x2,&rec[i].y2);        printf("Case %d:\n",++t);        memset(s,0,sizeof(s));        for ( i = 1; i <= m; i++ )        {            scanf("%d",&R);            cnt = add = 0; //用位运算记录那些被涂色的矩形            while ( R-- )            {                scanf("%d",&id);                if ( add & (1<<(id-1)) ) continue;                add = (add | (1<<(id-1)));                a[cnt++] = id;            }            if ( s[add] == 0 )                s[add] = In_Exclusion (0,total);            printf("Query %d: %d\n",i,s[add]);        }        printf("\n");    }    return 0;}