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Rectangles
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 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). 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 |
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。。。）
1 #include<bits/stdc++.h>
2 using namespace std;
3 struct node
4 {
5 int X1,Y1,X2,Y2;
6 }R[26];
7 int s1,cc[26],ans[26];
8 int read()
9 {
10 int s=0,fh=1;char ch=getchar();
11 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)fh=-1;ch=getchar();}
12 while(ch>=‘0‘&&ch<=‘9‘){s=s*10+(ch-‘0‘);ch=getchar();}
13 return s*fh;
14 }
15 void Cover(int X1,int Y1,int X2,int Y2,int k,int k1)
16 {
17 while(k<=s1&&(X1>=R[cc[k]].X2||X2<=R[cc[k]].X1||Y1>=R[cc[k]].Y2||Y2<=R[cc[k]].Y1))k++;
18 if(k>=s1+1){ans[k1]+=(X2-X1)*(Y2-Y1);return;}
19 if(X1<R[cc[k]].X1){Cover(X1,Y1,R[cc[k]].X1,Y2,k+1,k1);X1=R[cc[k]].X1;}
20 if(X2>R[cc[k]].X2){Cover(R[cc[k]].X2,Y1,X2,Y2,k+1,k1);X2=R[cc[k]].X2;}
21 if(Y1<R[cc[k]].Y1){Cover(X1,Y1,X2,R[cc[k]].Y1,k+1,k1);Y1=R[cc[k]].Y1;}
22 if(Y2>R[cc[k]].Y2){Cover(X1,R[cc[k]].Y2,X2,Y2,k+1,k1);Y2=R[cc[k]].Y2;}
23 }
24 int main()
25 {
26 int n,m,i,case1=0,C=0,j,sum;
27 while(1)
28 {
29 n=read();m=read();if(n==0&&m==0)break;
30 for(i=1;i<=n;i++){R[i].X1=read();R[i].Y1=read();R[i].X2=read();R[i].Y2=read();}
31 printf("Case %d:\n",++case1);
32 C=0;
33 for(i=1;i<=m;i++)
34 {
35 s1=read();
36 for(j=1;j<=s1;j++)cc[j]=read();
37 for(j=s1;j>=1;j--)Cover(R[cc[j]].X1,R[cc[j]].Y1,R[cc[j]].X2,R[cc[j]].Y2,j+1,j);//矩形切割法
38 sum=0;
39 for(j=1;j<=s1;j++){sum+=ans[j];ans[j]=0;}
40 printf("Query %d: %d\n",++C,sum);
41 }
42 printf("\n");
43 }
44 fclose(stdin);
45 fclose(stdout);
46 return 0;
47 }