HDU 5289 Assignment （ST算法区间最值+二分）

题目链接：

pid=5289">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.csdn.net/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;
}