POJ 1836 Alignment 枚举中间点双向求LIS

Description

Input

Output

Sample Input

8
1.86 1.86 1.30621 2 1.4 1 1.97 2.2

Sample Output

4

Source

//172K	297MS
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
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<n;i++)//枚举中间点
        {
            //求1到i的最长上升子序列
            dp[1]=tail[1];
            int low,high,mid,len=1;
            for(int j=2;j<=i;j++)
            {
                low=1;high=len;
                while(low<=high)
                {
                    mid=(low+high)>>1;
                    if(dp[mid]<tail[j])low=mid+1;
                    else high=mid-1;
                }
                dp[low]=tail[j];
                if(low>len)len++;
            }

            //求i+1到n的最长下降子序列
            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++)//倒着求最长上升子序列
            {
                low=1;high=len2;
                while(low<=high)
                {
                    mid=(low+high)>>1;
                    if(dp[mid]<s[j])low=mid+1;
                    else high=mid-1;
                }
                dp[low]=s[j];
                if(low>len2)len2++;
            }
            if(len+len2>maxx)maxx=len+len2;
        }
        printf("%d\n",n-maxx);
    }
    return 0;
}