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POJ 1836-Alignment(dp)

2024-11-25 10:58:02   203人阅读

Alignment
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 13624 Accepted: 4392

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：删除最少的人数，使得a1 < a2 < a3 < ... < a(i ) <=> a(i+1) > a(i+2) > .. a(n-1) >a(n)。

思路：求出从左到右的最长上升子序列，求出从右到左的最长上升子序列，然后用n-相加的和，注意这里有个坑，就是出现中间值为1个  或者中间值为2个的情况。例如1 2 3 2 1  和 1 2 3 3 2 1.这两种情况要分别考虑。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>

using namespace std;
const int inf=0x3f3f3f3f;
int dp1[1010];
int dp2[1010];
double a[1010];
int main()
{
    int n,i,j;
    while(~scanf("%d",&n)){
        for(i=0;i<n;i++)
            scanf("%lf",&a[i]);
        for(i=0;i<n;i++){
            dp1[i]=dp2[i]=1;
        }
        for(i=1;i<n;i++)
            for(j=0;j<i;j++)
            if(a[i]>a[j]&&dp1[i]<dp1[j]+1)
            dp1[i]=dp1[j]+1;
        for(i=n-2;i>=0;i--)
            for(j=n-1;j>i;j--)
            if(a[i]>a[j]&&dp2[i]<dp2[j]+1)
            dp2[i]=dp2[j]+1;
        int maxx=-inf;
        for(i=0;i<n;i++)
            maxx=max(maxx,dp1[i]+dp2[i]-1);//中间值为1个的时候
        for(i=0;i<n;i++)
            for(j=i+1;j<n;j++)
            maxx=max(maxx,dp1[i]+dp2[j]);//中间值为2个的时候
        printf("%d\n",n-maxx);
    }
    return 0;
}


POJ 1836-Alignment(dp)


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