HDOJ 题目5289 Assignment（RMQ，技巧）

Assignment

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1870    Accepted Submission(s): 916

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

field=problem&key=2015+Multi-University+Training+Contest+1&source=1&searchmode=source">2015 Multi-University Training Contest 1

 

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 题目大意：输入n,k,问n个数的序列。有多少个区间，最大值减最小值小于k

暴力找区间超时一次

ac代码，跑了1000ms

#include<stdio.h>
#include<string.h>
#include<math.h>
#define max(a,b) (a>b?a:b)
#define min(a,b) (a>b?b:a)
int n;
__int64 m;
__int64 a[100010];
__int64 maxv[100010][20],minv[100010][20];
void init()
{
	int i,j,k;
	for(i=1;i<=n;i++)
	{
		minv[i][0]=a[i];
		maxv[i][0]=a[i];
	}
	for(j=1;(1<<j)<=n;j++)
	{
		for(k=0;k+(1<<j)-1<=n;k++)
		{
			minv[k][j]=min(minv[k][j-1],minv[k+(1<<(j-1))][j-1]);
			maxv[k][j]=max(maxv[k][j-1],maxv[k+(1<<(j-1))][j-1]);
		}
	}
}
__int64 q_max(int l,int r)
{
	int k=(int)(log((double)(r-l+1))/log(2.0));
	return max(maxv[l][k],maxv[r-(1<<k)+1][k]);
}
__int64 q_min(int l,int r)
{
	int k=(int)(log((double)(r-l+1))/log(2.0));
	return min(minv[l][k],minv[r-(1<<k)+1][k]);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%I64d",&n,&m);
		int i,j,k;
		for(i=1;i<=n;i++)
		{
			scanf("%I64d",&a[i]);
		}
		init();
		__int64 ans=0;
		int p=1;
		/*for(i=1;i<=n;i++)//暴力超时
		{
			for(j=i;j<=n;j++)
			{
				if(q_max(i,j)-q_min(i,j)<m)
					ans++;
			}
		}*/
		for(i=1;i<=n;i++)
		{
			while(q_max(p,i)-q_min(p,i)>=m&&p<i)
			{
				p++;
			}
			ans+=i-p+1;
		}
		printf("%I64d\n",ans);
	}
}

网上看到用deque做的，跑了577ms，收藏一下，http://www.bubuko.com/infodetail-987302.html

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std ;
#define LL __int64
deque <LL> deq1 , deq2 ;
//单调队列，deq1最大值，deq2最小值
LL a[100010] ;
int main() {
    int t , n , i , j ;
    LL k , ans ;
    scanf("%d", &t) ;
    while( t-- ) {
        scanf("%d %I64d", &n, &k) ;
        for(i = 0 ; i < n ; i++)
            scanf("%I64d", &a[i]) ;
        if(k == 0) {
            printf("0\n") ;
            continue ;
        }
        while( !deq1.empty() ) deq1.pop_back() ;
        while( !deq2.empty() ) deq2.pop_back() ;
        for(i = 0 , j = 0 , ans = 0; i < n ; i++) {//i在前。j在后
            while( !deq1.empty() && deq1.back() < a[i] ) deq1.pop_back() ;
            deq1.push_back(a[i]) ;
            while( !deq2.empty() && deq2.back() > a[i] ) deq2.pop_back() ;
            deq2.push_back(a[i]) ;
            while( !deq1.empty() && !deq2.empty() && deq1.front() - deq2.front() >= k ) {
                ans += (i-j) ;
                //printf("%d %d,%I64d %I64d\n", i , j, deq1.front() , deq2.front() ) ;
                if( deq1.front() == a[j] ) deq1.pop_front() ;
                if( deq2.front() == a[j] ) deq2.pop_front() ;
                j++ ;
            }
        }
        while( j < n ) {
            ans += (i-j) ;
            j++ ;
        }
        printf("%I64d\n", ans) ;
    }
    return 0 ;
}