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+50;
const int maxm = 1e4+10;
const int inf = 0x3f3f3f3f;
const int mod = 998244353;
const double epx = 1e-10;
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=0;
    for(int i=x;i>0;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=1;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        memset(c,0,sizeof(c));
        memset(rankk,0,sizeof(rankk));
        int temp=2*n+m;
        for(int i=1;i<=temp;i++)
        {
            scanf("%d",&a[i].x);
            a[i].id=i;
        }
        sort(a+1,a+1+temp);
        int k=1;
        rankk[a[1].id]=k;
        for(int i=2;i<=temp;i++)
        {
            if(a[i].x==a[i-1].x)
                rankk[a[i].id]=k;
            else
                rankk[a[i].id]=++k;
        }
        for(int i=1;i<2*n;i+=2)
        {
            update(rankk[i],k,1);
            update(rankk[i+1]+1,k,-1);
        }
        printf("Case #%d:\n",kase++);
        for(int i=2*n+1;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。。