Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3995 Accepted Submission(s): 1308
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
题意：求一个最长子序列，使得最大值与最小值差值在mi，mx范围内，
用一个递增，一个递减两个单调队列维护最大值，最小值，然后扫描一遍，队列的起点从最近一个被单调队列抛弃的下标+1算起。
代码：
/* ***********************************************
Author :_rabbit
Created Time :2014/5/13 10:30:48
File Name :C.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int a[100100],que1[100100],que2[100100];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int n,mx,mi;
while(~scanf("%d%d%d",&n,&mi,&mx)){
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
int head1=0,tail1=0,head2=0,tail2=0;
int last1=0,last2=0;
int ans=0;
for(int i=1;i<=n;i++){
while(head1<tail1&&a[que1[tail1-1]]>a[i])tail1--;//递增
que1[tail1++]=i;
while(head2<tail2&&a[que2[tail2-1]]<a[i])tail2--;//递减
que2[tail2++]=i;
while(a[que2[head2]]-a[que1[head1]]>mx){
if(que1[head1]<que2[head2])head1++;
else head2++;
}
if(a[que2[head2]]-a[que1[head1]]>=mi)
ans=max(ans,i-max(que1[head1-1],que2[head2-1]));
}
cout<<ans<<endl;
}
return 0;
}
HDU 3530 单调队列