POJ 1836 Alignment

最新推荐文章于 2024-10-01 18:52:56 发布

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最新推荐文章于 2024-10-01 18:52:56 发布

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分类专栏：动态规划文章标签：动态规划

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解决一个有趣的动态规划问题，目标是最小化需要出列的士兵数量，以确保剩余士兵能够直接看到队列的一端。

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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

题意：

令到原队列的最少士兵出列后，使得新队列任意一个士兵都能看到左边或者右边的无穷远处。也就是说，严格递增、严格递减和先增后减都符合题意

题解：

很有意思的一个动态规划的思想，先把严格递增需要修改的人数列出，再把严格递减需要修改的人数列出，通过动态规划找出一个人作为最高点使解最优

CODE：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
    int dp1[1001],dp2[1001];
    double f[1001];
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1; i<=n; i++)
            scanf("%lf",&f[i]);
        for(int i=1; i<=n; i++)
            dp1[i]=dp2[i]=1;
        for(int i=1; i<=n; i++)
            for(int j=1; j<i; j++)
                if(f[i]>f[j])
                    dp1[i]=max(dp1[i],dp1[j]+1);
        for(int i=n; i>=1; i--)
            for(int j=n; j>i; j--)
                if(f[i]>f[j])
                    dp2[i]=max(dp2[i],dp2[j]+1);
        int ans=-1;
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                ans=max(ans,dp1[i]+dp2[j]);
        printf("%d\n",n-ans);
    }
    return 0;
}