Sticks Problem

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

题目大概：

是看一下给出的数列中是否存在单调递增的子序列，然后求子序列的长度-1。没有则是-1。

思路：

这个题用单调队列直接维护一个单调队列，然后求维护中最长的一个队列是多少就行了，但是tle，原来还要用别的方法，线段树后者其他，这个都没学，看到暴力破解可以，就用了暴力。

代码：

#include <iostream>#include <cmath>#include <cstdio>#include <cstring>using namespace std;int a[55000];int n;int main(){    while(cin>>n)    {        int sun=0;        int mid;        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }         int i=0;        while(i<n-1)      {          int sum=a[i];           mid=i;          int j;          for(j=i+1;j<n;j++)          {              if(a[j]<=a[i])              {                                    i=mid+1;                  break;              }              else if(a[j]<=sum)              {                  if(j==n-1)i=mid+1;                  continue;              }              else              {                  if(j-i>sun)sun=j-i;                  sum=a[j];                  mid=j;              }              if(j==n-1)i=mid+1;          }      }      if(sun==0)printf("%d\n",-1);      else printf("%d\n",sun);    }    return 0;}