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+];
 void ope(int x,int val)
 {
      x+=;
     while(x<=maxn)
     {
         aa[x]+=val;
         x+=lowbit(x);
     }
 }
 long long getsum(int x)
 {
    long long ans=;
     while(x>)
     {
         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,,sizeof(aa));
       for(i=;i<nn;i++)
       {
        scanf("%d%d",&a,&b);
        ope(a,);
        ope(b+,-);
       }
       for(i=;i<m;i++)
       {
        scanf("%d",&a);
        a+=;
        printf("%I64d\n",getsum(a));
       }
     }
   return ;
 }