Hdu 5289-Assignment 贪心,ST表

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

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3175    Accepted Submission(s): 1457

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

 

Sample Output

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 

 

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题意：给你一个数列，再给你一个k，问存在多少个连续子序列使得子序列的最大最小值差值 小于 k.

题解：

贪心＋ST表

枚举右端点，因为左端点一定是递增或不变的，所以遇到枚举的区间内的最大最小值差值 大于等于 k,就将左端点加1。然后每次统计个数即可。

注意：答案要开long long。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010
int n,mn[MAXN][17],mx[MAXN][17],a[MAXN];
int read()
{
    int s=0,fh=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')fh=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){s=s*10+(ch-'0');ch=getchar();}
    return s*fh;
}
void ST()
{
    int i,j;
    for(i=1;i<=n;i++)mn[i][0]=mx[i][0]=a[i];
    for(j=1;(1<<j)<=n;j++)
    {
        for(i=1;i+(1<<j)-1<=n;i++)
        {
            mn[i][j]=min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);
            mx[i][j]=max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);
        }
    }
}
int Query(int l,int r)
{
    int j;
    for(j=0;(1<<j)<=(r-l+1);j++);j--;
    return max(mx[l][j],mx[r-(1<<j)+1][j])-min(mn[l][j],mn[r-(1<<j)+1][j]);
}
int main()
{
    int T,k,i,left,right;
    long long ans;
    T=read();
    while(T--)
    {
        n=read();k=read();
        for(i=1;i<=n;i++)a[i]=read();
        ST();
        left=1;//左端点(左端点一定是单调递增的)
        ans=0;
        for(right=1;right<=n;right++)//枚举右端点
        {
            while(Query(left,right)>=k&&left<right)left++;//在枚举右端点的同时,移动左端点.每次改变右端点时,左端点只可能不变或向右移动.
            ans+=((long long)right-left+1LL);//统计的子串为以右端点为最右端,最左端在 右端点->左端点 之间.
        }
        printf("%lld
",ans);
    }
    fclose(stdin);
    fclose(stdout);
    return 0;
}