nyoj----522 Interval （简单树状数组）

Interval

时间限制：2000 ms  |  内存限制：65535 KB

难度：4

 

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

上传者

ACM_赵铭浩

 

代码：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define maxn 200005   //整体平移100001个单位
#define lowbit(x) ((x)&(-x))
int aa[maxn+5];
void ope(int x,int val)
{
     x+=100001;
    while(x<=maxn)
    {
        aa[x]+=val;
        x+=lowbit(x);
    }
}
long long getsum(int x)
{
   long long ans=0;
    while(x>0)
    {
        ans+=aa[x];
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    int test,nn,m,i,a,b;
    scanf("%d",&test);
    while(test--)
    {
      scanf("%d%d",&nn,&m);
      memset(aa,0,sizeof(aa));
      for(i=0;i<nn;i++)
      {
       scanf("%d%d",&a,&b);
       ope(a,1);
       ope(b+1,-1);
      }
      for(i=0;i<m;i++)
      {
       scanf("%d",&a);
       a+=100001;
       printf("%I64d
",getsum(a));
      }
    }
  return 0;
}