POJ 1836 Alignment 枚舉中間點雙向求LIS

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Alignment Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 13590 Accepted: 4375

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

一排士兵按照標號站成一排，讓一些士兵出列，使得剩下的士兵朝左或者朝右看都能看到無窮遠處。 where each soldier can see by looking lengthwise the line at least one of the line's extremity (left or right).士兵能夠朝著至少一個方向看到無窮遠處。
新隊列的身高要滿足這個式子：tail[1]tail[i+2]>......tail[n] 把從1枚舉到n-1，分別求出1到i的最長上升子序列和i+1到n的最長下降子序列，最後用n減去他們的最大和。

//172K	297MS
#include
#include
#include
#include
using namespace std;
double tail[1007],dp[1007],s[1007];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int maxx=-1;
        for(int i=1;i<=n;i++)
            scanf("%lf",&tail[i]);

        memset(dp,0,sizeof(dp));
        for(int i=1;i>1;
                    if(dp[mid]len)len++;
            }

            //求i+1到n的最長下降子序列
            memset(dp,0,sizeof(dp));
            memset(s,0,sizeof(s));
            int k=0,len2=1;
            for(int j=n;j>i;j--)//倒序一下
                s[++k]=tail[j];
            dp[1]=s[1];
            for(int j=2;j<=k;j++)//倒著求最長上升子序列
            {
                low=1;high=len2;
                while(low<=high)
                {
                    mid=(low+high)>>1;
                    if(dp[mid]len2)len2++;
            }
            if(len+len2>maxx)maxx=len+len2;
        }
        printf("%d\n",n-maxx);
    }
    return 0;
}