Sticks Problem--（单调队列）

Sticks Problem

Time Limit : 12000/6000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 64   Accepted Submission(s) : 19

Problem Description

Xuanxuan has n sticks of different length. One day, she puts all her sticks in a line, represented by S1, S2, S3, ...Sn. After measuring the length of each stick Sk (1 <= k <= n), she finds that for some sticks Si and Sj (1<= i < j <= n), each stick placed between Si and Sj is longer than Si but shorter than Sj. 

Now given the length of S1, S2, S3, …Sn, you are required to find the maximum value j - i.

 

Input

The input contains multiple test cases. Each case contains two lines. <br>Line 1: a single integer n (n <= 50000), indicating the number of sticks. <br>Line 2: n different positive integers (not larger than 100000), indicating the length of each stick in order.

 

Output

Output the maximum value j - i in a single line. If there is no such i and j, just output -1.

 

Sample Input

45 4 3 646 5 4 3

 

Sample Output

1-1

i和j之间数k(i<k<=j)满足2个条件：1.a[i]<a[k], 2.a[k]<=a[j];  ，很搞笑对吗，后面这两个条件是分别证的

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long int ll;int n;int a[50005],q[50005],r[50005];int main(){    int i,j;    while(scanf("%d",&n)!=EOF)    {        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        int tail=1; q[1]=n+1;        for(i=n;i>=1;i--)  //用单调队列求出a[i]右边连续大于他的数有几个        {            while(tail>=1&&a[q[tail]]>a[i]) tail--;            r[i]=q[tail]-1-i;            q[++tail]=i;        }        int maxn,temp,ans=0;        for(i=1;i<=n;i+=temp+1)        {            maxn=temp=-1;            for(j=0;j<=r[i];j++) //对于大于他的这些数，都是满足条件1的            {                if(a[i+j]>maxn) //只要a[i+j]大于maxn，说明他是目前最大的也就是以他结尾的话前面a[i+1]至a[i+j]都满足条件2                {                    maxn=a[i+j];                    temp=j;  //此时记录下j，下面i+=temp+1;就从那些没算到的开始                }            }            if(ans<temp) ans=temp;        }        if(ans!=0) printf("%d\n",ans);        else printf("-1\n");    }    return 0;}