poj-1836-士兵出列问题-双向LIS

Alignment

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 17175 Accepted: 5644

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

Source

Romania OI 2002

题意：让一列士兵中出列最少的士兵数，使剩下的士兵存在从左到右或从右到左（二者可同时存在）的高度递增序列

方法：

#include <iostream>#include <algorithm>using namespace std;int num1[1010];int num2[1010];double a[1010];int main(){    int n;    while(cin >> n)    {        for(int i=0; i<n; i++)            cin >> a[i];        for(int i=0; i<n; i++)            num1[i] = 1;        for(int i=n-2; i>=0; i--)//O(n2)方法可以记录每个数据左右的递增序列长度        {            for(int j=n-1; j>i; j--)            {                if(a[i]>a[j] && num1[i]<num1[j]+1)                    num1[i] = num1[j]+1;            }        }        for(int i=0; i<n; i++)            num2[i] = 1;        for(int i=1; i<n; i++)        {            for(int j=0; j<i; j++)            {                if(a[i]>a[j] && num2[i]<num2[j]+1)                    num2[i] = num2[j]+1;            }        }        int mx = 0;        for(int i=0; i<n; i++)//当只有一个峰顶时            mx = max(mx, num1[i]+num2[i]-1);        for(int i=0; i<n; i++)//当有两个相邻的峰顶时        {            for(int j=i+1; j<n; j++)                mx = max(mx, num2[i]+num1[j]);        }        cout << n-mx << endl;    }    return 0;}