poj 3695 Rectangles（矩形切割）

Rectangles Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 3112   Accepted: 864 Description You are developing a software for painting rectangles on the screen. The software supports drawing several rectangles and filling some of them with a color different from the color of the background. You are to implement an important function. The function answer such queries as what is the colored area if a subset of rectangles on the screen are filled. Input The input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 20) and M(1 ≤ M ≤ 100000), indicating the number of rectangles on the screen and the number of queries, respectively. The i-th line of the following N lines contains four integers X1,Y1,X2,Y2 (0 ≤ X1 < X2 ≤ 1000, 0 ≤ Y1 < Y2 ≤ 1000), which indicate that the lower-left and upper-right coordinates of the i-th rectangle are (X1, Y1) and (X2,Y2). Rectangles are numbered from 1 to N. The last M lines of each test case describe M queries. Each query starts with a integer R(1<=R ≤ N), which is the number of rectangles the query is supposed to fill. The following list of R integers in the same line gives the rectangles the query is supposed to fill, each integer of which will be between 1 and N, inclusive. The last test case is followed by a line containing two zeros. Output For each test case, print a line containing the test case number( beginning with 1). For each query in the input, print a line containing the query number (beginning with 1) followed by the corresponding answer for the query. Print a blank line after the output for each test case. Sample Input 2 2 0 0 2 2 1 1 3 3 1 1 2 1 2 2 1 0 1 1 2 2 1 3 2 2 1 2 0 0 Sample Output Case 1: Query 1: 4 Query 2: 7 Case 2: Query 1: 2 Source 2008 Asia Hefei Regional Contest Online by USTC

题目：http://poj.org/problem?id=3695

题意：在一个平面上画矩形，给你n个矩形，每次从里面挑出若干个，问着几个矩形覆盖的面积，提问次数比较多

分析：这题提问数据比较多，不过矩行数比较少，个人感觉直接矩形切割就行了，所以写了个，居然1000ms过了，而且前20，刚好在ACRush前一名，离大神如此的近啊，膜拜中。。。这题的另一个做法就是线段树，不过最近太懒，不想写了= =

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int mm=22;
int sx[mm],sy[mm],tx[mm],ty[mm],d[mm];
int n,m,r,sum;
void Cut(int i,int ax,int ay,int bx,int by)
{
    while(i<r&&(ax>=tx[d[i]]||bx<=sx[d[i]]||ay>=ty[d[i]]||by<=sy[d[i]]))++i;
    if(i>=r)sum+=(bx-ax)*(by-ay);
    else
    {
        if(ax<sx[d[i]])Cut(i+1,ax,ay,sx[d[i]],by),ax=sx[d[i]];
        if(bx>tx[d[i]])Cut(i+1,tx[d[i]],ay,bx,by),bx=tx[d[i]];
        if(ay<sy[d[i]])Cut(i+1,ax,ay,bx,sy[d[i]]);
        if(by>ty[d[i]])Cut(i+1,ax,ty[d[i]],bx,by);
    }
}
int main()
{
    int i,cs=0,t;
    while(scanf("%d%d",&n,&m),n+m)
    {
        for(i=1;i<=n;++i)
            scanf("%d%d%d%d",&sx[i],&sy[i],&tx[i],&ty[i]);
        printf("Case %d:\n",++cs);
        t=0;
        while(m--)
        {
            scanf("%d",&r);
            for(i=0;i<r;++i)
                scanf("%d",&d[i]);
            for(sum=i=0;i<r;++i)
                Cut(i+1,sx[d[i]],sy[d[i]],tx[d[i]],ty[d[i]]);
            printf("Query %d: %d\n",++t,sum);
        }
        puts("");
    }
    return 0;
}