Poj 3695-Rectangles 矩形切割

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 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
 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(X11,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(Y11,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 }

 

转载于:https://www.cnblogs.com/Var123/p/5377508.html