poj 3695 Rectangles（矩形切割）_boost用线段分割矩形-优快云博客

该博客主要介绍了POJ 3695题目的解题思路，题目要求在平面上处理多个矩形，并计算特定矩形组合的覆盖面积。作者采用了矩形切割的方法，虽然数据量较大，但通过这种方法实现了1000ms内的解决方案，且在比赛中取得了较好的成绩。另一种可能的解法是使用线段树，但由于作者的惰性并未实现。附有代码。

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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

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=369 5

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;
}