HDU4325 树状数组+离散化

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4325

Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3544    Accepted Submission(s): 1745

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  n朵花 m个时间点 第i朵花正在开放的时间是si，ti   问：在第mi个时间点有多少朵花正在开放

解析  思路是 我们记录每个时间点有多少个花是开放的   树状数组可以解决 但是数据太大会超时也存不下 数据数量还是比较少的 所以我们可以离散化一下 我们要把m也离散化进去

AC代码

 #include <stdio.h>
 #include <math.h>
 #include <string.h>
 #include <stdlib.h>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 #include <iomanip>
 using namespace std;
 const int maxn = 3e5+;
 const int maxm = 1e4+;
 const int inf = 0x3f3f3f3f;
 const int mod = ;
 const double epx = 1e-;
 typedef long long ll;
 const ll INF = 1e18;
 const double pi = acos(-1.0);
 int c[maxn],rankk[maxn];
 int n,m,t;
 struct node
 {
     int id,x;
     bool operator<(const node &b)const
     {
         return x<b.x;
     }
 }a[maxn];
 int lowbit(int x)
 {
     return x&(-x);
 }
 int getsum(int x)
 {
     int ans=;
     for(int i=x;i>;i-=lowbit(i))
         ans+=c[i];
     return ans;
 }
 void update(int x,int y,int z)
 {
     for(int i=x;i<=y;i+=lowbit(i))
         c[i]+=z;
 }
 int main()
 {
     int kase=;
     cin>>t;
     while(t--)
     {
         cin>>n>>m;
         memset(c,,sizeof(c));
         memset(rankk,,sizeof(rankk));
         int temp=*n+m;
         for(int i=;i<=temp;i++)
         {
             scanf("%d",&a[i].x);
             a[i].id=i;
         }
         sort(a+,a++temp);
         int k=;
         rankk[a[].id]=k;
         for(int i=;i<=temp;i++)
         {
             if(a[i].x==a[i-].x)
                 rankk[a[i].id]=k;
             else
                 rankk[a[i].id]=++k;
         }
         for(int i=;i<*n;i+=)
         {
             update(rankk[i],k,);
             update(rankk[i+]+,k,-);
         }
         printf("Case #%d:\n",kase++);
         for(int i=*n+;i<=temp;i++)
         {
             printf("%d\n",getsum(rankk[i]));
         }
     }
 }

离散化博客    https://blog.csdn.net/xiangaccepted/article/details/73276826

数据比较水 不离散化也可以水过去。。