NYOJ-522 Interval

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

      很久没写树状数组了，正好看到了，就复习一下，本题是树状数组-----插线问点的问题

注意写树状数组时候把区间都扩大到大于0的区间上，不能等于0.否则会一直超时，代码如下：

 

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

int num[200005];

int lowbit(int i)
{
    return i & (-i);
}

void update(int i, int add)
{
    while(i <= 200001)
    {
        num[i] += add;
        i += lowbit(i);
    }
}

int getsum(int x)
{
    int ans = 0;
    while(x > 0)
    {
        ans += num[x];
        x -= lowbit(x);
    }
    return ans;
}

int main()
{
    int T, m, n, s, t, x;
    scanf("%d", &T);
    while(T--)
    {
        memset(num, 0, sizeof(num));
        scanf("%d%d", &m, &n);
        while(m--)
        {
            scanf("%d%d", &s, &t);
            update(s+100001, 1);
            update(t+100001+1, -1);
        }
        while(n--)
        {
            scanf("%d", &x);
            printf("%d\n", getsum(x+100001));
        }
    }
    return 0;
}