描述

There are n(1 <= n <= 100000) intervals [ai, bi] and m(1 <= m <= 100000) queries, -100000 <= ai <= bi <= 100000 are integers.

Each query contains an integer xi(-100000 <= x <= 100000). For each query, you should answer how many intervals convers xi.

输入

The first line of input is the number of test case.
For each test case,
two integers n m on the first line, 
then n lines, each line contains two integers ai, bi;
then m lines, each line contains an integer xi.

输出

m lines, each line an integer, the number of intervals that covers xi.

样例输入

2
3 4
1 3
1 2
2 3
0
1
2
3
1 3
0 0
-1
0
1

样例输出

0
2
3
2
0
1
0

/*树状数组应用，统计数字在所给区间出现的次数。
TLE哭了，MAX大小：200002是AC，200005是TLE！！！
注意将负数转换到正值区间上，否则还是会TLE！！！ 
*/
#include<stdio.h>
#include<string.h>
#define MAX 200002
int a[MAX];
int lowbit(int N)
{
	return N&(-N);
}
void asd(int i,int M)
{
	while(i<=MAX)
	{
		a[i]+=M;
		i+=lowbit(i);
	}
}
int sum(int j)
{
	int sum=0;
	while(j>0)
	{
		sum+=a[j];
		j-=lowbit(j);
	}
	return sum;
}
int main()
{
	int i,n,m,T,t,k,p;
	scanf("%d",&T);
	while(T--)
	{
		memset(a,0,sizeof(a));
		scanf("%d%d",&m,&n);
		while(m--)
		{
			scanf("%d%d",&k,&t);
			asd(k+100001,1);
			asd(t+100002,-1);
		}
		while(n--)
		{
			scanf("%d",&p);
			printf("%d\n",sum(p+100001));
		}
	}
	return 0;
}

Interval（南阳oj522）（树状数组）