hdu3530 单调队列

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5809    Accepted Submission(s): 1911

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

 

 

题意:

求n个数里面最长的一段，其最大值减最小值的差>=m && <= k。

思路:

维护2个单调队列。一个存最大值，一个存最小值。

 

/*
 * Author:  sweat123
 * Created Time:  2016/7/12 9:09:45
 * File Name: main.cpp
 */
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = ;
int a[MAXN];
deque<int>q1,q2;
int n,m,k;
int main(){
    while(~scanf("%d%d%d",&n,&m,&k)){
        q1.clear();
        q2.clear();
        for(int i = ; i <= n; i++){
            scanf("%d",&a[i]);
        }
        int ans = ;
        int bf = ;
        for(int i = ; i <= n; i++){
            while(!q1.empty() && a[q1.back()] < a[i]){
                q1.pop_back();
            }
            while(!q2.empty() && a[q2.back()] > a[i]){
                q2.pop_back();
            }
            q1.push_back(i);
            q2.push_back(i);
            while(!q1.empty() && !q2.empty() && a[q1.front()] - a[q2.front()] > k){
                if(q1.front() < q2.front()){
                    bf = q1.front();
                    q1.pop_front();
                } else if(q1.front() > q2.front()){
                    bf = q2.front();
                    q2.pop_front();
                } else {
                    bf = q1.front();
                    q1.pop_front();
                    q2.pop_front();
                }
            }
            if(!q1.empty() && !q2.empty() && a[q1.front()] - a[q2.front()] >= m){
                ans = max(ans,i - bf);
            }
        }
        printf("%d\n",ans);
    }
    return ;
}