hdu5289||2015多校聯合第一場1002貪心+RMQ

life4711發表於2015-07-21
MQ

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

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] 
 





 



Source



2015 Multi-University Training Contest 1





/**
hdu5289||2015多校聯合第一場1002貪心+RMQ
題目大意：找出連續子區間，其最大最小值只差小於k
解題思路：用rmq維護區間最大最小值只差，貪心掃一遍即可，複雜度O(nlogn)
*/
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int N=200005;
int a[N],n,m;
int dp1[N][30];
int dp2[N][30];

void RMQ_init(int n)
{
    for(int i=1;i<=n;i++)
    {
        dp1[i][0]=a[i];
        dp2[i][0]=a[i];
    }
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i+(1<<j)-1<=n;i++)///白書上的模板，次行稍作改動，否則dp陣列要擴大一倍防止RE
        {
            dp1[i][j]=max(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);
            dp2[i][j]=min(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);
        }
    }
}

int rmq(int x,int y)
{
    int k=0;
    while((1<<(k+1))<=y-x+1)k++;
    return max(dp1[x][k],dp1[y-(1<<k)+1][k])-min(dp2[x][k],dp2[y-(1<<k)+1][k]);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        RMQ_init(n);
        int k=1;
        long long sum=0;
        for(int i=1;i<=n;i++)
        {
            while(rmq(k,i)>=m&&k<i)k++;
            sum+=(i-k+1);
           // printf("dis i k i-k+1:%d %d %d %d\n",rmq(k,i),i,k,i-k+1);
        }
        printf("%I64d\n",sum);
    }
    return 0;
}








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