SERC2013 J You Win!

装压DP......

http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11450

J:   You Win!
You just  achieved the High  Score on your favorite video game! Now, you get to enter 
your name! You have to use the  controller  to enter  your name, which can be awkward. 
Here’s how it works:
  There  are  only  the  26  capital  letters  A  to  Z,  in  order.  There  are  no  numbers, 
spaces, lower case letters, or any other characters.
  Pushing  UP  or  DOWN  changes  the  active  letter  one  letter  forward  (UP)  or 
backward  (DOWN).  The active  letter  starts at  A.  It will not reset  when you move 
around in the name. It also wraps: UP from Z goes to A, DOWN from A goes to Z.
  Pushing  LEFT  or  RIGHT  moves  the  cursor  one  letter  left  or  right  in  the  current 
name. Note that once the cursor is at either end of the current name, it cannot 
move any further in that direction.
  Pushing the FIRE button adds the active letter to the name. 
For example, consider the name ‘ALMA’. One way you could enter ‘ALMA’ is like this:
Action  # of Pushes  Name (| = Cursor)  Active Letter
FIRE  1  A|  A
UP  11  A|  L
FIRE  1  AL|  L
UP  1  AL|  M
FIRE  1  ALM|  M
DOWN  12  ALM|  A
FIRE  1  ALMA|  A
This would take 28 button pushes. However, consider entering ‘ALMA’ like this:
Action  # of Pushes  Name (| = Cursor)  Active Letter
FIRE  1  A|  A
FIRE  1  AA|  A
LEFT  1  A|A  A
UP  11  A|A  L
FIRE  1  AL|A  L
UP  1  AL|A  M
FIRE  1  ALM|A  M 
2013 ACM ICPC Southeast USA Regional Programming Contest
Page 16 of 16    2 November, 2013
This takes only 17  button pushes. Given a  name, what is the fewest number of  button 
pushes needed to enter that name? Assume that the active  letter  starts at A, and that it 
doesn’t matter where the cursor ends up when you’re done.
Input
There will be  several  test cases in the input. Each test  case will  consist of a single string 
on  its  own  line,  with  from  1  to  18  capital  letters,  representing  a  name  that  must  be 
entered into the High Score list. The input will end with a line with a single 0.
Output
For each test case, output a single integer representing the smallest number of  button 
pushes needed to enter the name. Output no spaces, and do not separate answers with 
blank lines.
Sample Input
ALMA
YES
0
Sample

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int INF=0x3f3f3f3f;

int dp[1<<19][19],n;
char str[20];

int getD(char a,char c)
{
    if(a>c) swap(a,c);
    return min(c-a,a-'A'+'Z'-c+1);
}

int getSHIFT(int from,int to,int st)
{
    int ans=0;
    if(from<to)
    {
        for(int i=from;i<to;i++)
        {
            if(st&(1<<i)) ans++;
        }
    }
    else
    {
        for(int i=from-1;i>to-1;i--)
        {
            if(st&(1<<i)) ans++;
        }
    }
    return ans;
}
void solve()
{
    memset(dp,63,sizeof(dp));
    for(int i=0;i<19;i++) dp[0][i]=0;
    for(int i=0;i<n;i++)
    {
        dp[1<<i][i+1]=getD('A',str[i])+1;
    }
    for(int i=1;i<(1<<n);i++)
    {
        for(int j=1;j<=n;j++)
        {
            ///在状态i  光标在j
            ///PS1: 除状态0外光标必须在哪个字符后面,光标前面的字符即上一个字母,也就是从没有j-1号字母的
            ///状态转移过来的
            if((i&(1<<(j-1)))==0) continue;
            if(dp[i][j]!=INF) continue;
            int k=j-1;
            int ii=i^(1<<k);///从没有k号字符的状态开始转移 即状态ii-->状态i
            for(int jj=1;jj<=n;jj++)///枚举状态ii的光标位置
            {
                if((ii&(1<<(jj-1)))==0) continue;
                dp[i][j]=min(dp[i][j],dp[ii][jj]+getD(str[jj-1],str[j-1])+getSHIFT(jj,j,ii)+1);
            }
        }
    }
    int ans=INF;
    for(int i=1;i<=n;i++) ans=min(ans,dp[(1<<n)-1][i]);
    printf("%d\n",ans);
}


int main()
{
while(scanf("%s",str)!=EOF)
{
    if(str[0]=='0') break;
    n=strlen(str);
    solve();
}
    return 0;
}