hdu2461 Rectangles-----容斥

Rectangles
Time Limit: 5000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 989    Accepted Submission(s): 558

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

Source
2008 Asia Hefei Regional Contest Online by USTC

Recommend
teddy
题意：给n个矩形，然后每次涂m个矩形，问涂上色的面积是多少。
利用容斥原理即可。
[cpp]
#include<iostream> 
#include<cstdlib> 
#include<stdio.h> 
#include<algorithm> 
using namespace std; 
int n,m,num,res; 
int a[25]; 
struct Node 
{ 
    int x1,y1,x2,y2; 
    int s; 
}node[25]; 
bool check(Node x,int u) 
{ 
    if(node[u].x1>=x.x2) return 0; 
    if(x.x1>=node[u].x2) return 0; 
    if(node[u].y1>=x.y2) return 0; 
    if(x.y1>=node[u].y2) return 0; 
    return 1; 
} 
void solve(Node rec,int id,int k) 
{ 
   Node temp; 
   if(id==num+1) return ; 
   for(int i=id;i<=num;i++) 
   { 
       if(check(rec,a[i])) 
       { 
         temp.x1=rec.x1>node[a[i]].x1?rec.x1:node[a[i]].x1; 
         temp.y1=rec.y1>node[a[i]].y1?rec.y1:node[a[i]].y1; 
         temp.x2=rec.x2<node[a[i]].x2?rec.x2:node[a[i]].x2; 
         temp.y2=rec.y2<node[a[i]].y2?rec.y2:node[a[i]].y2; 
         temp.s=(temp.x2-temp.x1)*(temp.y2-temp.y1); 
         if(k%2) res+=temp.s; 
         else res-=temp.s; 
         solve(temp,i+1,k+1); 
       } 
   } 
} 
int main() 
{ 
    int count=1; 
    while(scanf("%d%d",&n,&m)!=EOF) 
    { 
        if(n==0&&m==0) break; 
        for(int i=1;i<=n;i++) 
        { 
            scanf("%d%d%d%d",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2); 
            node[i].s=(node[i].x2-node[i].x1)*(node[i].y2-node[i].y1); 
        } 
        printf("Case %d:/n",count++); 
        for(int i=1;i<=m;i++) 
        { 
            scanf("%d",&num); 
            res=0; 
            for(int j=1;j<=num;j++) 
            { 
                scanf("%d",&a[j]); 
                res+=node[a[j]].s; 
            } 
            for(int j=1;j<=num;j++) 
            { 
                solve(node[a[j]],j+1,0); 
            } 
            printf("Query %d: ",i); 
            printf("%d/n",res); 
        } 
        puts(""); 
    } 
    return 0; 
}