## locker(dp) HDU - 4433

A password locker with N digits, each digit can be rotated to 0-9 circularly.
You can rotate 1-3 consecutive digits up or down in one step.
For examples:
567890 -> 567901 (by rotating the last 3 digits up)
000000 -> 000900 (by rotating the 4th digit down)
Given the current state and the secret password, what is the minimum amount of steps you have to rotate the locker in order to get from current state to the secret password?

InputMultiple (less than 50) cases, process to EOF.
For each case, two strings with equal length (≤ 1000) consists of only digits are given, representing the current state and the secret password, respectively.
OutputFor each case, output one integer, the minimum amount of steps from the current state to the secret password.

Sample Input

```111111 222222
896521 183995```

Sample Output

```2
12```

：上升枚举或者下降枚举，另外，注意在状态转移的过程中,第2个位置转换次数要小于第1个位置的转换次数。

AC代码：
``` 1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<algorithm>
5 using namespace std;
6 const int maxn = 1e3+10;
7 #define LL long long
8 #define INF 0x3f3f3f3f
9 char s1[maxn],s2[maxn];
10 int a[maxn],b[maxn];
11 int dp[maxn][10][10];
12
13
14 int main(){
15     int i,j,k,p,q,n,m,len,up,down;
16     while(~scanf("%s%s",s1+1,s2+1)){
17         len=strlen(s1+1);
18         a[0]=b[0]=0;
19         a[len+1]=a[len+2]=0;
20         b[len+1]=b[len+2]=0;
21         for(int i=1;i<=len;i++){
22             a[i]=s1[i]-'0';
23             b[i]=s2[i]-'0';
24         }
25         memset(dp,INF,sizeof(dp));
26         dp[0][a[1]][a[2]]=0;
27         for(int i=1;i<=len;i++){
28             for(int j=0;j<10;j++){
29                 for(int k=0;k<10;k++){
30                     if(dp[i-1][j][k]==INF) continue;
31                     down=(b[i]-j+10)%10;
32                     for(p=0;p<=down;p++) for(q=0;q<=p;q++){
33                         dp[i][(k+p)%10][(a[i+2]+q)%10]=min(dp[i][(k+p)%10][(a[i+2]+q)%10],dp[i-1][j][k]+down);
34                     }
35                     up=10-down;
36                     for(p=0;p<=up;p++) for(q=0;q<=p;q++){
37                         dp[i][(k-p+10)%10][(a[i+2]-q+10)%10]=min(dp[i][(k-p+10)%10][(a[i+2]-q+10)%10],dp[i-1][j][k]+up);
38                     }
39                 }
40             }
41         }
42         printf("%d\n",dp[len][0][0]);
43     }
44
45
46
47     return 0;
48 }
49 /*
50 111111 222222
51 896521 183995
52 */```

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