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POJ 1836-Alignment（DP/LIS變形）

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POJ 1836-Alignment（DP/LIS變形）

Alignment Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 13465 Accepted: 4336

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

最長上升子序列變形；

題意： 一排士兵，要求對於每一個士兵，至少保證左邊或者右邊的人身高單調遞增。問滿足要求最少要出去多少人。這個問題等價於 最多有多少個士兵可以按要求排成一列。掃兩遍LIS ，枚舉每一個點 求 max(dp_l[i]+dp_r[j]);i:1 to n; j:i+1 to n;

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define maxn 116
#define pp pair
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n,dp_l[1010],dp_r[1010];
double a[1010];
void solve()
{
	for(int i=2;i<=n;i++)
		for(int j=1;ja[j]&&dp_l[i]<=dp_l[j])
				dp_l[i]=dp_l[j]+1;
	for(int i=n-1;i>=1;i--)
		for(int j=n;j>i;j--)
			if(a[i]>a[j]&&dp_r[i]<=dp_r[j])
				dp_r[i]=dp_r[j]+1;
	int ans=-INF;
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
		ans=max(ans,dp_l[i]+dp_r[j]);
	printf("%d\n",n-ans);
}
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=1;i<=n;i++)
		{
			dp_l[i]=1;
			dp_r[i]=1;
			scanf("%lf",&a[i]);
		}
		solve();
	}
	return 0;
}