Poj 3695-Rectangles 矩形切割

2024-03-21 14:44:52

Rectangles

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3846		Accepted: 1124

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

题意：给你一些矩形，编号为1~n。有一些询问，每次问你其中的一些矩形的面积并。

题解：

矩形切割

矩形切割裸题。。。当然也可以用线段树（大坑。。。）

直接挂个模版就过了。

看到Discuss中有人说矩形切割会TLE。可能是常数太大了吧。但我没有TLE呀，而且貌似跑的飞快（时间好像排32名，内存也很小。。。）

 #include<bits/stdc++.h>

 using namespace std;

 struct node

 {

     int X1,Y1,X2,Y2;

 }R[];

 int s1,cc[],ans[];

 int read()

 {

     int s=,fh=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')fh=-;ch=getchar();}

     while(ch>=''&&ch<=''){s=s*+(ch-'');ch=getchar();}

     return s*fh;

 }

 void Cover(int X1,int Y1,int X2,int Y2,int k,int k1)

 {

     while(k<=s1&&(X1>=R[cc[k]].X2||X2<=R[cc[k]].X1||Y1>=R[cc[k]].Y2||Y2<=R[cc[k]].Y1))k++;

     if(k>=s1+){ans[k1]+=(X2-X1)*(Y2-Y1);return;}

     if(X1<R[cc[k]].X1){Cover(X1,Y1,R[cc[k]].X1,Y2,k+,k1);X1=R[cc[k]].X1;}

     if(X2>R[cc[k]].X2){Cover(R[cc[k]].X2,Y1,X2,Y2,k+,k1);X2=R[cc[k]].X2;}

     if(Y1<R[cc[k]].Y1){Cover(X1,Y1,X2,R[cc[k]].Y1,k+,k1);Y1=R[cc[k]].Y1;}

     if(Y2>R[cc[k]].Y2){Cover(X1,R[cc[k]].Y2,X2,Y2,k+,k1);Y2=R[cc[k]].Y2;}

 }

 int main()

 {

     int n,m,i,case1=,C=,j,sum;

     while()

     {

         n=read();m=read();if(n==&&m==)break;

         for(i=;i<=n;i++){R[i].X1=read();R[i].Y1=read();R[i].X2=read();R[i].Y2=read();}

         printf("Case %d:\n",++case1);

         C=;

         for(i=;i<=m;i++)

         {

             s1=read();

             for(j=;j<=s1;j++)cc[j]=read();

             for(j=s1;j>=;j--)Cover(R[cc[j]].X1,R[cc[j]].Y1,R[cc[j]].X2,R[cc[j]].Y2,j+,j);//矩形切割法

             sum=;

             for(j=;j<=s1;j++){sum+=ans[j];ans[j]=;}

             printf("Query %d: %d\n",++C,sum);

         }

         printf("\n");

     }

     fclose(stdin);

     fclose(stdout);

     return ;

 }

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