Editing a Book UVA

You have

n

equal-length paragraphs numbered 1 to

n

. Now you want to arrange them in the order

of 1

;

2

;:::;n

. With the help of a clipboard, you can easily do this: Ctrl-X (cut) and Ctrl-V (paste)

several times. You cannot cut twice before pasting, but you can cut several contiguous paragraphs at

the same time - they'll be pasted in order.

For example, in order to make

f

2, 4, 1, 5, 3, 6

g

, you can cut 1 and paste before 2, then cut 3 and

paste before 4. As another example, one copy and paste is enough for

f

3, 4, 5, 1, 2

g

. There are two

ways to do so: cut

f

3, 4, 5

g

and paste after

f

1, 2

g

, or cut

f

1, 2

g

and paste before

f

3, 4, 5

g

.

Input

The input consists of at most 20 test cases. Each case begins with a line containing a single integer

n

(1

<n<

10), thenumber of paragraphs. The next line contains a permutation of 1

;

2

;

3

;:::;n

. The

last case is followed by a single zero, which should not be processed.

Output

For each test case, print the case number and the minimal number of cut/paste operations.

Sample Input

6

2 4 1 5 3 6

5

3 4 5 1 2

0

Sample Output

Case 1: 2

Case 2: 1

/**
题目：Editing a Book UVA - 11212
链接：https://vjudge.net/problem/UVA-11212
题意：lrjP208.
思路：启发函数：当前已经操作的步数d，限制的步数maxd，后继不相同的个数x。
由于每一步最多使x减少3，所以如果3*d+x>maxd*3;那么肯定不行。

如果通过位置不同的个数作为启发，是不行的，无法判断。

如果是每个数的后继不同的数有多少个。那么可以推算出每一步操作后，最多减少3个不同的后继。

最终结果所有后继都满足。即a[i] = a[i-1]+1;

具体看lrj算法竞赛入门经典P209；


*/

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long LL;
const int mod=1e9+7;
const int maxn=1e2+5;
const double eps = 1e-12;
int a[10], n;
int getDif()///获得后继不合法数.
{
    int cnt = 0;
    for(int i = 2; i <= n; i++){
        if(a[i]!=a[i-1]+1){
            cnt++;
        }
    }
    return cnt;
}
int temp[10];
void Move(int *temp,int *a,int L,int R,int LL,int RR)
{
    int now = L;
    for(int i = LL; i <= RR; i++){
        a[now++] = temp[i];
    }
    for(int i = L; i <= R; i++){
        a[now++] = temp[i];
    }
}
bool dfs(int d,int maxd)
{
    if(3*d+getDif()>3*maxd) return false;
    if(getDif()==0) return true;
    int odd[10];
    memcpy(odd,a,sizeof(int)*(n+1));
    ///如果[L,R]是连续的，那么不需要剪切。
    for(int i = 1; i <= n; i++){
        for(int j = i; j <= n; j++){
            for(int k = j+1; k <= n; k++){///每次交换[i,j],[j+1,k]；每次交换相邻的两段区间。
                Move(odd,a,i,j,j+1,k);
                if(dfs(d+1,maxd)) return true;
                memcpy(a,odd,sizeof(int)*(n+1));///保证每一次操作后，把a复原。
            }
        }
    }
    return false;
}
int main()
{
    int cas = 1;
    while(scanf("%d",&n)==1)
    {
        if(n==0) break;
        for(int i = 1; i <= n; i++) scanf("%d",&a[i]);
        for(int maxd = 0; ; maxd++){
            if(dfs(0, maxd)){
                printf("Case %d: %d
",cas++,maxd); break;
            }
        }
    }
    return 0;
}