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Alignment
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: Output
The only line of output will contain the number of the soldiers who have to get out of the line.
Sample Input 8 Sample Output 4 Source |
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最近在进行动态规划的专题练习。。。
题意：令到原队列的最少士兵出列后，使得新队列任意一个士兵都能看到左边或者右边的无穷远处。
可以是递增序列。也可以是递减序列。更可以是如下图：
所以这是双向lis问题，刚开始的时候还没想到是lis问题，就按照自己的方法做。结果wa。后来发现有很多bug。。
看了discuss才想到是lis问题。还是练得少呀！！
#include<cstdio>
#include<cstring>
int dp1[1006]; //dp1[i]代表以i结尾的最长上升子序列
int dp2[1006]; //dp2[i]代表以i开始的最长下降子序列
int maxdp[1006]; //maxdp[i]代表以0~i中dp1中的最最大值
double p[1006];
int Max(int a,int b)
{
return a>b?a:b;
}
int main()
{
int n,i,j,mm;
while(scanf("%d",&n)!=EOF)
{
mm = 0;
for(i=1;i<=n;i++)
scanf("%lf",&p[i]);
maxdp[1] = dp1[1] = 1;
maxdp[0] = 0;
for(i=2;i<=n;i++) //两个lis算法
{
dp1[i] = 1;
for(j=1;j<i;j++)
{
if(p[j]<p[i])
dp1[i] = Max(dp1[i],dp1[j]+1); }
maxdp[i] = Max(maxdp[i-1],dp1[i]);
}
dp2[n] = 1;
for(i=n-1;i>=1;i--)
{
dp2[i] = 1;
for(j=n;j>i;j--)
{
if(p[j]<p[i])
dp2[i] = Max(dp2[i],dp2[j]+1); }
mm = Max(mm,maxdp[i]+dp2[i+1]); //把以i前面的上升序列最大值和以i+1结尾的最长下降子序
//列长度加起来和前面对比，这样就是把全部的情况都算了一遍
} printf("%d\n",n-mm);
}
return 0;
}