Rectangles hdu2461容斥定理

Rectangles

Time Limit: 5000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1259    Accepted Submission(s): 661

Problem 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

 

 

没优化版：

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <algorithm>
 using namespace std;
 #define INF 100000000
 typedef struct point
 {
     int x1,y1,x2,y2;
 }point;
 point p[];
 int ans[]={};
 int n;
 void dfs(int x1,int y1,int x2,int y2,int deep,int sign,int sta)
 {
     if( x1 >= x2 || y1 >= y2 ) return;
     if(deep==n)
     {
         if(sta)
         for(int i=;i<(<<n);i++)
         {
             if((i|sta)<=i)
             ans[i]+=sign*(x2-x1)*(y2-y1);
         }
         return ;
     }
     dfs(x1,y1,x2,y2,deep+,sign,sta);
     dfs(max(x1,p[deep].x1),max(y1,p[deep].y1),min(x2,p[deep].x2),min(y2,p[deep].y2),deep+,-sign,sta|(<<deep));
 }
 int main()
 {
     int m,i,ss,cas=,mm,x,cass;
     while(scanf("%d%d",&n,&m),(n||m))
     {
         memset(ans,,sizeof(ans));
         for(i=;i<n;i++)
             scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
         dfs(,,INF,INF,,-,);
         printf("Case %d:\n",cas++);
         cass=;
         while(m--)
         {
             scanf("%d",&mm);
             ss=;
             for(i=;i<mm;i++)
             {
                 scanf("%d",&x);
                 ss|=(<<(x-));
             }
             printf("Query %d: %d\n",cass++,ans[ss]);
         }
         printf("\n");
     }
 }

优化版：

 #include <iostream>
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <algorithm>
 using namespace std;
 #define INF 100000000
 typedef struct point
 {
     int x1,y1,x2,y2;
 } point;
 point p[];
 int ans[],staa[];
 int n,m;
 void dfs(int x1,int y1,int x2,int y2,int deep,int sign,int sta)
 {
     if( x1 >= x2 || y1 >= y2 ) return;
     if(deep==n)
     {
         if(sta)
             for(int i=; i<m; i++)
             {
                 if((staa[i]|sta)<=staa[i])
                     ans[staa[i]]+=sign*(x2-x1)*(y2-y1);
             }
         return ;
     }
     dfs(x1,y1,x2,y2,deep+,sign,sta);
     dfs(max(x1,p[deep].x1),max(y1,p[deep].y1),min(x2,p[deep].x2),min(y2,p[deep].y2),deep+,-sign,sta|(<<deep));
 }
 int main()
 {
     int i,cas=,mm,x,cass;
     while(scanf("%d%d",&n,&m),(n||m))
     {
         memset(ans,,sizeof(ans));
         memset(staa,,sizeof(staa));
         for(i=; i<n; i++)
             scanf("%d%d%d%d",&p[i].x1,&p[i].y1,&p[i].x2,&p[i].y2);
         printf("Case %d:\n",cas++);
         cass=;
         while(m--)
         {
             scanf("%d",&mm);
             for(i=; i<mm; i++)
             {
                 scanf("%d",&x);
                 staa[cass]|=(<<(x-));
             }
             cass++;
         }
         m=cass;
         dfs(,,INF,INF,,-,);
         for(i=; i<=cass; i++)
             printf("Query %d: %d\n",i,ans[staa[i-]]);
         printf("\n");
     }
 }