hdu3530 Subsequence 单调队列

Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4903    Accepted Submission(s): 1615

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 01 1 1 1 15 0 31 2 3 4 5

 

Sample Output

54

  问最长的一个子序列长度是多少，子序列满足最大值减去最小值在m和k之间。

  用两个队列，最大值队列和最小值队列，最大值队列单调递减，最小值队列单调递增，最大值队列里的点u表示从u开始到当前点的最大值是a[u]，最小值队列里的点u表示从u开始到当前点的最小值是a[u]。一旦最大值队列首元素的值和最小值队列首元素的值之差不大于k，就判断这两个值的差是否不小于m，是就更新答案，注意更新答案的长度是从两个队列队首上个点加1里面大的那个值到当前点这段长度。如果最大值队列首元素的值和最小值队列首元素的值之差大于k，就把靠前的那个front加1，再继续判断。要注意细节问题，我为了方便处理先在队首加了个权值INF的点。

  用单调队列处理后复杂度是O（N）。

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;const LL MAXN=100010;const LL MAXM=5010;const LL INF=0x3f3f3f3f;int N,M,K;int a[MAXN],q1[MAXN],q2[MAXN];int main(){    freopen("in.txt","r",stdin);    while(scanf("%d%d%d",&N,&M,&K)!=EOF){        for(int i=1;i<=N;i++) scanf("%d",&a[i]);        a[0]=INF;        int front1=0,rear1=0,front2=0,rear2=0;        q1[0]=q2[0]=0;        int ans=0;        for(int i=1;i<=N;i++){            while(front1<=rear1&&a[q1[rear1]]<=a[i]) rear1--;            q1[++rear1]=i;            while(front2<=rear2&&a[q2[rear2]]>=a[i]) rear2--;            q2[++rear2]=i;            while(a[q1[front1]]-a[q2[front2]]>K){                if(q1[front1]<q2[front2]) front1++;                else front2++;            }            if(a[q1[front1]]-a[q2[front2]]>=M){                if(front2<=0) ans=max(ans,i-q1[front1-1]);                ans=max(ans,i-max(q1[front1-1],q2[front2-1]));            }        }        printf("%d\n",ans);    }    return 0;}