Flowers（树状数组+区间更新+单点查询+区间更新单点查询模板）

Problem I

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 30   Accepted Submission(s) : 18

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[sub]i[/sub] and T[sub]i[/sub] (1 <= S[sub]i[/sub] <= T[sub]i[/sub] <= 10^9), means i-th flower will be blooming at time [S[sub]i[/sub], T[sub]i[/sub]]. In the next M lines, each line contains an integer T[sub]i[/sub], 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

21 15 1042 31 44 8146

 

Sample Output

Case #1:0Case #2:121

题意：

每种花有开放的时间段，然后给你某一个时间，求这一时间内的开花数量。

思路：

简单的树状数组，只需要区间更新单点查询即可，套模板就行

代码：

#include <iostream>#include <cstring>#include <stdio.h>#include <algorithm>const int maxn=1e5+10;using namespace std;int a[maxn];int n,m;int lowbit(int i){return i&-i;}int sum(int i){int summ=0;while(i>0){summ+=a[i];i-=lowbit(i);}return summ;}void add(int i,int x){while(i<=100005){a[i]+=x;i+=lowbit(i);}}int main(){int t;int i,j;int s,e;scanf("%d",&t);for(int ss=1;ss<=t;ss++){memset(a,0,sizeof(a));printf("Case #%d:\n",ss);scanf("%d%d",&n,&m);while(n--){scanf("%d%d",&s,&e);add(s,1);add(e+1,-1);}while(m--){scanf("%d",&j);printf("%d\n",sum(j));}}return 0;}