Interval（南阳oj522）（树状数组）

Interval

2000 ms  |  内存限制： 65535

4

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.

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;
}