HDU 5289 Assignment （ST算法区间最值+二分）

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
(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
HintFirst 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，怎么想都想不对，想去找最近的不合法的点，复杂度太高。看了题解才知道是用ST算法的。先前不知道，这是一篇很不错的ST算法的介绍。/article/1944874.html

枚举左边端点，二分右端点，用ST算法判断该区间是否合法，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;
}