Alignment
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 13319 | Accepted: 4282 |
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
题目要求：给出n个人排成一排，踢出一些人，让每个人都能看到最左端，或最后端，最小的踢出人数是？
计算出正序和倒序的最长上升子序列，然后统计：有两种可能，一种是其中一个人是中间，那个人的身高最高，还有事两个人的身高相同，这两个人位置在中间，统计出最长的可能出现的队伍长度，计算出最小的踢出人数
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp1[1200] , dp2[1200] ;
double h[1200] ;
int main()
{
int i , j , n , min1 ;
while(scanf("%d", &n)!=EOF)
{
min1 = 1200 ;
h[0] = 0 ; h[n+1] = 0 ;
for(i = 1 ; i <= n ; i++)
scanf("%lf", &h[i]);
memset(dp1,0,sizeof(dp1));
for(i = 1 ; i <= n ; i++)
{
for(j = 0 ; j < i ; j++)
if( h[j] < h[i] && dp1[j]+1 > dp1[i] )
dp1[i] = dp1[j]+1 ;
}
memset(dp2,0,sizeof(dp2));
for(i = n ; i >= 1 ; i--)
{
for(j = n+1 ; j > i ; j--)
if( h[j] < h[i] && dp2[j]+1 > dp2[i] )
dp2[i] = dp2[j]+1 ;
}
for(i = 1 ; i <= n ; i++)
{
for(j = i ; j <= n ; j++)
{
if(i == j)
min1 = min(min1,n-(dp1[i]+dp2[j]-1) );
else
min1 = min(min1, n-( dp1[i]+dp2[j] ) );
}
}
printf("%d\n", min1);
}
}