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
題意：給n個矩形，然後每次塗m個矩形，問塗上色的面積是多少。
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;
}
