2012 #3 Flowers

Flowers

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4325

Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

 

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases. 
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times. 
In the next N lines, each line contains two integer S _i and T _i (1 <= S _i <= T _i <= 10^9), means i-th flower will be blooming at time [S _i, T _i].
In the next M lines, each line contains an integer T _i, means the time of i-th query. 

 

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers. 
Sample outputs are available for more details. 

 

Sample Input

2 1 1 5 10 4 2 3 1 4 4 8 1 4 6

 

Sample Output

Case #1: 0 Case #2: 1 2 1

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
    int T;
    int i,j;
    int n,m,ca=1;
    int l[100005],r[100005];
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%d %d",&l[i],&r[i]);
        sort(l+1,l+n+1);
        sort(r+1,r+n+1);
        printf("Case #%d:
",ca);ca++;
        for(i=1;i<=m;i++)
        {
            int x,y,z;
            scanf("%d",&z);
            x=upper_bound(l+1,l+n+1,z)-(l+1);
            y=upper_bound(r+1,r+n+1,z-1)-(r+1);
            printf("%d
",x-y);
        }
    }
    return 0;
}