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：

 1 /*树状数组模板题，若在区间a,b内开花则从a开始向上+1，从b开始向上-1，最后输出
 2 某一点的值即可。*/
 3 #include<iostream>
 4 #include<string>
 5 #include<cstdio>
 6 #include<cmath>
 7 #include<cstring>
 8 #include<algorithm>
 9 #include<vector>
10 #include<iomanip>
11 #include<queue>
12 #include<stack>
13 using namespace std;
14 int t,n,m;
15 int A[100005];
16 int lowbit(int x)
17 {
18     return x&(-x);
19 }
20 void add(int rt,int c)
21 {
22     while(rt<=100005)
23     {
24         A[rt]+=c;
25         rt+=lowbit(rt);
26     }
27 }
28 int sum(int rt)
29 {
30     int s=0;
31     while(rt>0)
32     {
33         s+=A[rt];
34         rt-=lowbit(rt);
35     }
36     return s;
37 }
38 int main()
39 {
40     int a,b,c;
41     scanf("%d",&t);
42     for(int i=1;i<=t;i++)
43     {
44         memset(A,0,sizeof(A));
45         printf("Case #%d:\n",i);
46         scanf("%d%d",&n,&m);
47         while(n--)
48         {
49             scanf("%d%d",&a,&b);
50             add(a,1);
51             add(b+1,-1);
52         }
53         while(m--)
54         {
55             scanf("%d",&c);
56             printf("%d\n",sum(c));
57         }
58     }
59     return 0;
60 }