HDU4325 树状数组

Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3148    Accepted Submission(s): 1549

Problem Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

 

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times.
In the next N lines, each line contains two integer S_i and T_i (1 <= S_i <= T_i <= 10^9), means i-th flower will be blooming at time [S_i, T_i].
In the next M lines, each line contains an integer T_i, means the time of i-th query.

 

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

 

Sample Input

2 1 1 5 10 4 2 3 1 4 4 8 1 4 6

 

Sample Output

Case #1: 0 Case #2: 1 2 1

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 题意：

给出若干个花的开花时期，问某一个日期有几多花开放。

代码：

/*树状数组模板题，若在区间a,b内开花则使a端点+1，使b+1端点-1，最后求到某一点的和就是某一点的值
某一点的值即可。*/
#include<iostream>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<queue>
#include<stack>
using namespace std;
int t,n,m;
int A[100005];
int lowbit(int x)
{
    return x&(-x);
}
void add(int rt,int c)
{
    while(rt<=100005)
    {
        A[rt]+=c;
        rt+=lowbit(rt);
    }
}
int sum(int rt)
{
    int s=0;
    while(rt>0)
    {
        s+=A[rt];
        rt-=lowbit(rt);
    }
    return s;
}
int main()
{
    int a,b,c;
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        memset(A,0,sizeof(A));
        printf("Case #%d:
",i);
        scanf("%d%d",&n,&m);
        while(n--)
        {
            scanf("%d%d",&a,&b);
            add(a,1);
            add(b+1,-1);
        }
        while(m--)
        {
            scanf("%d",&c);
            printf("%d
",sum(c));
        }
    }
    return 0;
}