POJ - 1836 Alignment

Alignment

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16494		Accepted: 5406

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

Source

Romania OI 2002

题意是从原序列中移走几个士兵，得到一个新的序列，且这个序列满足  每个士兵向左或向右至少一个方向都不存在比他高或和他一样高的

要求最少移走几个人

从前到后求一遍最长递增子序列，在从后到前求一遍

因为数据不大，最后两层for循环遍历求出最大值

又错在了忘记了dp【i】并非前i个的最长递增子序列长度，而应该是dp【1~i】里的最大值，所以最后只用了一层for循环遍历，谨记！

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
typedef long long ll;
using namespace std;
int main()
{	
	int n,dp[2000],dp1[2000],i,j,ans;
	double k[2000];
	scanf("%d",&n);
	for(i=0;i<n;i++)
	scanf("%lf",&k[i]);
	for(i=0;i<n;i++)
	dp[i]=1,dp1[i]=1;
	for(i=1;i<n;i++)
	{
		for(j=0;j<i;j++)
		{
			if(k[i]>k[j])
			dp[i]=max(dp[i],dp[j]+1);
		}
	}
	for(i=n-1;i>=0;i--)
	{
		for(j=n-1;j>i;j--)
		{
			if(k[i]>k[j])
			dp1[i]=max(dp1[i],dp1[j]+1);
		}
	}
	ans=dp[0]+dp1[0]-1;
	for(i=0;i<n;i++)
	for(j=i+1;j<n;j++)
	ans=max(ans,dp[i]+dp1[j]);
	
	for(i=1;i<n;i++)
	ans=max(ans,dp[i]+dp1[j]-1);
	printf("%d\n",n-ans);
	return 0;
 }