Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3150    Accepted Submission(s): 1549

Problem 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

Author

BJTU

Source

2012 Multi-University Training Contest 3

/*
给你每一朵话的开花时间段，询问你某一时刻的开花数量
*/

/*
重新定义树状数组的意义，不再是前i个数的和，而是第i个位置的数值
*/
/*
明显数据会爆的，我去.....数据太水了
*/
#include<iostream>
#include<string.h>
#include<stdio.h>
#define N 100010
using namespace std;
int c[N],T[N];
int t;
int n,m;
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int val)
{
    while(x<=N)
    {
        c[x]+=val;
        x+=lowbit(x);
    }
}
int getsum(int x)
{
    int s=0;
    while(x>0)
    {
        s+=c[x];
        x-=lowbit(x);
    }
    return s;
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    scanf("%d",&t);
    for(int Case=1;Case<=t;Case++)
    {
        memset(c,0,sizeof c);
        scanf("%d%d",&n,&m);
        //cout<<"n="<<n<<" "<<"m="<<m<<endl;
        int si,ti;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&si,&ti);
            //cout<<si<<" "<<ti<<endl;

            update(si,1);

            // for(int j=si;j<=ti;j++)
                // update(j,1);
            update(ti+1,-1);
        }
        for(int i=0;i<m;i++)
            scanf("%d",&T[i]);
        printf("Case #%d:\n",Case);
        // for(int i=1;i<10;i++)
             // printf("%d\n",getsum(i));
        for(int i=0;i<m;i++)
        {
            //cout<<"T[i]="<<T[i]<<" "<<"T[i]-1="<<T[i]-1<<endl;
            printf("%d\n",getsum(T[i]));
        }
    }
}