Alignment
| Time Limit: 1000MS |
|
Memory Limit: 30000K |
| Total Submissions: 7642 |
|
Accepted: 2434 |
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
4
一些士兵站成一排，現在要盡量少的士兵出來，使得剩些的士兵都能看到排左或排右，看到的意思是，中間沒有比它高的
這個題和合唱隊形差不多，但是有區(qū)別，中間的兩個人可以一樣高
從左到右求最長上升子序列，再右到左求最長上升子序列，
然后枚舉中間節(jié)點，求兩個序列的最大和
中間兩個一樣高可以，需要特別處理下 1
#include<stdio.h>
2#include<string.h>
3#include<math.h>
4#define eps 0.0000001
5#define MAX 1005
6double a[MAX];
7int f[MAX],f1[MAX];
8int n,i,j,ans;
9int max(int a,int b)
10
{
11
if (a>b) return a;
12 else return b;
13
}
14int main()
15{
16 scanf("%d",&n);
17 for (i=1; i<=n ; i++ ) scanf("%lf",&a[i]);
18 f[1]=1;
19 for (i=2; i<=n ; i++ )
20
{
21 f[i]=1;
22 for (j=1; j<=i-1 ; j++ )
23 {
24 if ((a[i]-a[j])>eps)
25 {
26 f[i]=max(f[j]+1,f[i]);
27
}
28 }
29 }
30 f1[n]=1;
31 for (i=n-1; i>=1 ; i-- )
32 {
33 f1[i]=1;
34 for (j=n; j>=i+1 ; j-- )
35 {
36 if ((a[i]-a[j])>eps)
37 {
38 f1[i]=max(f1[i],f1[j]+1);
39 }
40 }
41 }
42 ans=0;
43 for (i=1; i<=n ; i++ )
44 {
45 ans=max(ans,f[i]+f1[i]-1);
46 for (j=i+1;j<=n ;j++ )
47 {
48 if ((a[i]-a[j])<eps)
49 {
50 if (ans<f[i]+f1[j])
51 {
52 ans=f[i]+f1[j];
53 }
54 else break;
55 }
56 }
57 }
58 printf("%d\n",n-ans);
59 return 0;
60}
61/**////合唱隊形類似
62