POJ 1836 Alignment(DP LIS)

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

81.86 1.86 1.30621 2 1.4 1 1.97 2.2

Sample Output

4

题目大意：在一个序列中最少删除多少人可以让队列中的所有人看到左无穷或者右无穷。

即形成一个三角形，或者中间有两个身高相同的梯形,etc..

思路：求删除最少的人就是队列中留下最多的人成了LIS问题，只需要，从左到右，然后再反过来一边，求解出dp[]与invertdp[]，然后在进行枚举0-i,i-n-1两部分

的正序LIS和逆序LIS的和，相加得到ans,最后n-ans;

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#define LL long long#define inf 0x3f3f3f3fusing namespace std;double a[1010];int dp[1010],invertdp[1010];int main(){    int n,m,x,k,i,j,cla;    while(~scanf("%d",&n))    {        for(i=0;i<n;i++)            scanf("%lf",&a[i]);        memset(dp,0,sizeof(dp));        memset(invertdp,0,sizeof(invertdp));        dp[0]=1;        for(i=1;i<n;i++)        {            dp[i]=1;            for(j=0;j<i;j++)                if(a[i]>a[j]&&dp[i]<dp[j]+1)                    dp[i]=dp[j]+1;        }        invertdp[n-1]=1;        for(i=n-2;i>=0;i--)        {            invertdp[i]=1;            for(j=n-1;j>i;j--)                if(a[i]>a[j]&&invertdp[i]<invertdp[j]+1)                    invertdp[i]=invertdp[j]+1;        }        int ans=0;        for(i=0;i<n-1;i++)        {            for(j=i+1;j<n;j++)            {                ans=max(ans,dp[i]+invertdp[j]);            }        }        printf("%d\n",n-ans);    }    return 0;}