poj1836——dp,最长上升子序列(lis)

Alignment

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 13767		Accepted: 4450

Description

In the army, a platoon is composed by n soldiers. During the morning inspection, the soldiers are aligned in a straight line in front of the captain. The captain is not satisfied with the way his soldiers are aligned; it is true that the soldiers are aligned in order by their code number: 1 , 2 , 3 , . . . , n , but they are not aligned by their height. The captain asks some soldiers to get out of the line, as the soldiers that remain in the line, without changing their places, but getting closer, to form a new line, where each soldier can see by looking lengthwise the line at least one of the line's extremity (left or right). A soldier see an extremity if there isn't any soldiers with a higher or equal height than his height between him and that extremity.

Write a program that, knowing the height of each soldier, determines the minimum number of soldiers which have to get out of line.

Input

On the first line of the input is written the number of the soldiers n. On the second line is written a series of n floating numbers with at most 5 digits precision and separated by a space character. The k-th number from this line represents the height of the soldier who has the code k (1 <= k <= n).

There are some restrictions:
• 2 <= n <= 1000
• the height are floating numbers from the interval [0.5, 2.5]

Output

The only line of output will contain the number of the soldiers who have to get out of the line.

Sample Input

8
1.86 1.86 1.30621 2 1.4 1 1.97 2.2

Sample Output

4

题意：使队列中出列部分士兵，使剩余士兵身高先单增后单减（单调部分不可相等)，求最少的出列士兵数量思路：求两次lis（最长上升子序列），再遍历得出答案；lis：dp(i)=max{dp(j)+1,dp(i)}，初始化dp[]={1};

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cstdio>using namespace std;const int maxn=1000100;
int n;
double h[maxn];
int dp1[maxn],dp2[maxn];int main()
{while(cin>>n){for(int i=0;i<n;i++){cin>>h[i];dp1[i]=dp2[i]=1;}for(int i=0;i<n;i++){  //求最长上升子序列(lis)for(int j=0;j<i;j++){if(h[j]<h[i]) dp1[i]=max(dp1[j]+1,dp1[i]);}}for(int i=n-1;i>=0;i--){for(int j=n-1;j>i;j--){if(h[j]<h[i]) dp2[i]=max(dp2[j]+1,dp2[i]);}}int ans=0;for(int i=0;i<n-1;i++){for(int j=i+1;j<n;j++){if(dp1[i]+dp2[j]>ans) ans=dp1[i]+dp2[j];}}cout<<n-ans<<endl;}return 0;
}

View Code

当然lis有nlogn的做法

转载于:https://www.cnblogs.com/--560/p/4352315.html

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