POJ 1836 - Alignment（最长递增子序列）

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本文探讨了一个在军队中优化士兵排列的问题，通过递增和递减的算法来减少不必要的移动，以确保每个士兵都能看到队伍的两端至少有一侧。文中详细介绍了输入输出格式、数据限制和解决方案。

Alignment

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11559		Accepted: 3684

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

==============================

求个递增求个递减，注意递增递减边界两只是可以等高的，不是严格先递增后递减，然后枚举

#include <iostream>
#include <cstring>
using namespace std;

double a[1111];
int n,ans;
int f[1111],g[1111];

int main()
{
    while(cin>>n)
    {
        memset(f,0,sizeof(f));
        memset(g,0,sizeof(g));
        for(int i=1; i<=n; i++)
        {
            cin>>a[i];
        }
        a[0]=-1e9;
        a[n+1]=-1e9;
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<i; j++)
            {
                if (a[i]>a[j]&&f[j]+1>f[i])
                {
                    f[i]=f[j]+1;
                }
            }
        }
        for(int i=n; i>=1; i--)
        {
            for(int j=n+1; j>i; j--)
            {
                if(a[i]>a[j]&&g[j]+1>g[i])
                {
                    g[i]=g[j]+1;
                }
            }
        }
        ans=0;
        for(int i=1; i<=n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
                ans=max(ans,f[i]+g[j]);
            }
        }
        ans=n-ans;
        cout<<ans<<endl;
    }
    return 0;
}