HDU 3530 Subsequence

Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 7550 Accepted Submission(s): 2555

Problem Description

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

Input

There are multiple test cases.

For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range [1, 100000]. m and k are in the range [0, 1000000]. The second line has n integers, which are all in the range [0, 1000000].

Proceed to the end of file.

Output

For each test case, print the length of the subsequence on a single line.

Sample Input

5 0 0

1 1 1 1 1

5 0 3

1 2 3 4 5

Sample Output

5

4

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

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//维护两个单调队列
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define N 100010
using namespace std;
inline int max(int a ,int b) { return a>b?a:b; }
int s1[N],s2[N],a[N];
int main() {
    int n,m,k,top1,top2,last1,last2,tail1,tail2,Ans;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF) {
        for(int i=; i<=n; i++) scanf("%d",&a[i]);
        memset(s1,,sizeof(s1));
        memset(s2,,sizeof(s2));
        top1 = ,top2 = ,tail1 = ,tail2 = ;
        Ans = ,last1 = ,last2 = ;
        for(int i=; i<=n; i++){
            //max
            while(top1 < tail1 && a[s1[tail1 - ]] <= a[i]) tail1--;  //top1最大元素
            s1[tail1++] = i;
            //min
            while(top2 < tail2 && a[s2[tail2 - ]] >= a[i]) tail2--;  //top2最小元素
            s2[tail2++] = i;
            while(a[s1[top1]] - a[s2[top2]] > k){
                if(s1[top1] < s2[top2])
                    last1 = s1[top1++];
                else last2 = s2[top2++];
            }
            if(a[s1[top1]] - a[s2[top2]] >= m)
                Ans = max(Ans,i - max(last1,last2));
        }
        printf("%d\n",Ans);
    }
    return ;
}