HDU 5289 Assignment （ST算法区间最值+二分）

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原文链接：http://www.cnblogs.com/blfbuaa/p/7106408.html

文章标签：

#java #数据结构与算法 #php

本文介绍了解决HDU 5289 Assignment问题的方法，使用ST算法来确定满足条件的员工组数。通过枚举左端点并二分查找右端点，判断区间内员工能力值之差是否符合要求。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5289

题面：

Assignment

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 672 Accepted Submission(s): 335

Problem Description

Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. Tom want to know the number of groups like this.

Input

In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1<=n<=100000, 0<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,a[n](0<=a[i]<=10^9),indicate the i-th staff’s ability.

Output

For each test，output the number of groups.

Sample Input

    2 4 2 3 1 2 4 10 5 0 3 4 5 2 1 6 7 8 9
   

    5 28 
    
      Hint
     
First Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]

Author

FZUACM

Source

2015 Multi-University Training Contest 1

解题：

比赛的时候，怎么想都想不正确。想去找近期的不合法的点，复杂度太高。

看了题解才知道是用ST算法的。先前不知道，这是一篇非常不错的ST算法的介绍。

http://blog.youkuaiyun.com/david_jett/article/details/46990651

枚举左边端点，二分右端点。用ST算法推断该区间是否合法，直至右端点到极限（即二分的左右边界相遇或交叉）。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include <vector>
#include <cmath>
#include <algorithm>
#define mod 1000000007
using namespace std;
int t,n,k;
int a[100100],minn[100010][20],maxn[100010][20],mid;
long long ans;
void Rmq_Init()
{  
    int m=19;  
    for(int i=1;i<=n;i++) 
        maxn[i][0]=minn[i][0]=a[i];  
    for(int i=1;i<=m;i++)  
        for(int j=n;j;j--)
        {  
            maxn[j][i]=maxn[j][i-1];  
            minn[j][i]=minn[j][i-1];  
            if(j+(1<<(i-1))<=n)
            {  
                maxn[j][i]=max(maxn[j][i],maxn[j+(1<<(i-1))][i-1]);  
                minn[j][i]=min(minn[j][i],minn[j+(1<<(i-1))][i-1]);  
            }  
        }  
}  
int Query_dif(int l,int r)
{  
    int m=floor(log((double)(r-l+1))/log(2.0));  
    int Max=max(maxn[l][m],maxn[r-(1<<m)+1][m]);  
    int Min=min(minn[l][m],minn[r-(1<<m)+1][m]);  
    return Max-Min;  
}  
int solve(int l)
{
    int le,ri;
    le=l;
    ri=n;
        while(le<=ri)
        {
            mid=(le+ri)/2;
            if(Query_dif(l,mid)>=k)
            {
                ri=mid-1;
            }
            else
            {
                le=mid+1;
            }
        }
        /*if(Query_dif(l,mid)>=k)
          return mid-l;
        else 
          return mid-l+1;*/
        return ri-l+1;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
      ans=0;
      scanf("%d%d",&n,&k);
      for(int i=1;i<=n;i++)
      {
          scanf("%d",&a[i]);
      }
      Rmq_Init();
      for(int i=1;i<=n;i++)
      {
        ans=ans+solve(i);
        //cout<<i<<" "<<Query_dif(i,n)<<endl;
        //cout<<ans<<endl;
      }
      printf("%lld\n",ans);
    }
    return 0;
}

转载于:https://www.cnblogs.com/blfbuaa/p/7106408.html