HDU 4433 locker

Problem Description

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?

 

Input

Multiple (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.

 

Output

For each case, output one integer, the minimum amount of steps from the current state to the secret password.

 

Sample Input

111111 222222896521 183995

 

Sample Output

212

 

f[now][i][j]表示从第1位到第now位全部已经对上且现在第now+1位是i，第now+2位是j的最少步数,然后递推转移即可

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1005;char s1[maxn],s2[maxn];int f[maxn][10][10],n,a[maxn],b[maxn],ans;void get(int a,int b,int c,int x){    if (f[a][b][c]<0) f[a][b][c]=x;    else f[a][b][c]=min(f[a][b][c],x);}int main(){    while (~scanf("%s%s",s1,s2))    {        n=strlen(s1);        ans=9*n;        for (int i=1;i<=n;i++) a[i]=s1[i-1]-'0',b[i]=s2[i-1]-'0';        a[n+1]=a[n+2]=b[n+1]=b[n+2]=0;        for (int i=0;i<=n;i++) memset(f[i],-1,sizeof(f[i]));        f[0][a[1]][a[2]]=0;        for (int now=0;now<n;now++)        {            int end=a[now+3];            for (int i=0;i<10;i++)            {                int up=(b[now+1]-i+10)%10;                int dw=(i-b[now+1]+10)%10;                for (int j=0;j<10;j++)                    if (f[now][i][j]>=0)                    {                        for (int k=0;k<=up;k++)                            for (int kk=0;kk<=k;kk++)                                get(now+1,(j+k)%10,(end+kk)%10,f[now][i][j]+up);                        for (int k=0;k<=dw;k++)                            for (int kk=0;kk<=k;kk++)                                get(now+1,(10+j-k)%10,(10+end-kk)%10,f[now][i][j]+dw);                    }            }        }        for (int i=0;i<10;i++)            for (int j=0;j<10;j++)                if (f[n][i][j]>=0) ans=min(f[n][i][j],ans);        printf("%d\n",ans);    }    return 0;}