Rectangles
Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 3927   Accepted: 1150

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


【题意】给N个矩形，他们可能会重叠，然后给M个询问，每个询问给出指定的矩形位置，然后分别计算每个询问中选中的矩形的并。

【解题方法】如上，直接粘模板让我WA了很多次啊，之后又TLE，看了讨论知道用C++可以过，果然1200ms过掉了。

【AC 代码】

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
struct node{
    LL x1,y1,x2,y2;
    LL sum;
    void read()
    {
        scanf("%I64d%I64d%I64d%I64d",&x1,&y1,&x2,&y2);
    }
}T[44];
LL n,m;
LL a[44];
LL xx;
void Cover(LL x1,LL y1,LL x2,LL y2,LL k,LL c)
{
    while(k<xx&&(x1>=T[a[k]].x2||x2<=T[a[k]].x1||y1>=T[a[k]].y2||y2<=T[a[k]].y1)) k++;
    if(k>=xx){
        T[a[c]].sum+=(x2-x1)*(y2-y1);
        return ;
    }
    if(x1<T[a[k]].x1){
        Cover(x1,y1,T[a[k]].x1,y2,k+1,c);
        x1=T[a[k]].x1;
    }
    if(x2>T[a[k]].x2){
        Cover(T[a[k]].x2,y1,x2,y2,k+1,c);
        x2=T[a[k]].x2;
    }
    if(y1<T[a[k]].y1){
        Cover(x1,y1,x2,T[a[k]].y1,k+1,c);
        y1=T[a[k]].y1;
    }
    if(y2>T[a[k]].y2){
        Cover(x1,T[a[k]].y2,x2,y2,k+1,c);
        y2=T[a[k]].y2;
    }
}
int main()
{
    LL ks1=1,ks2;
    while(scanf("%I64d%I64d",&n,&m)!=EOF)
    {
        if(n==0&&m==0) break;
        memset(T,0,sizeof(T));
        for(int i=0; i<n; i++) T[i].read();
        printf("Case %I64d:\n",ks1++);
        ks2=1;
        while(m--)
        {
            memset(a,0,sizeof(a));
            for(int i=0; i<n; i++) T[i].sum=0LL;
            scanf("%I64d",&xx);
            for(int i=0; i<xx; i++){
                scanf("%I64d",&a[i]);
                a[i]--;
            }
            for(LL i=xx-1; i>=0; i--){
                Cover(T[a[i]].x1,T[a[i]].y1,T[a[i]].x2,T[a[i]].y2,i+1,i);
            }
            LL ans=0;
            for(int i=0; i<xx; i++){
                ans+=T[a[i]].sum;
            }
            printf("Query %I64d: %I64d\n",ks2++,ans);
        }
        printf("\n");
    }
    return 0;
}