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[];
 int lowbit(int x)
 {
     return x&(-x);
 }
 void add(int rt,int c)
 {
     while(rt<=)
     {
         A[rt]+=c;
         rt+=lowbit(rt);
     }
 }
 int sum(int rt)
 {
     int s=;
     while(rt>)
     {
         s+=A[rt];
         rt-=lowbit(rt);
     }
     return s;
 }
 int main()
 {
     int a,b,c;
     scanf("%d",&t);
     for(int i=;i<=t;i++)
     {
         memset(A,,sizeof(A));
         printf("Case #%d:\n",i);
         scanf("%d%d",&n,&m);
         while(n--)
         {
             scanf("%d%d",&a,&b);
             add(a,);
             add(b+,-);
         }
         while(m--)
         {
             scanf("%d",&c);
             printf("%d\n",sum(c));
         }
     }
     return ;
 }