hdu4433 locker 密码锁（枚举）

locker

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 883    Accepted Submission(s): 374

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

 

Source

2012 Asia Tianjin Regional Contest

 

dp[i][j][k]表示 前i个已经完全匹配，而这时候，第i+1个已经加了j位，第i+2位已经加了k

转移分为两步，枚举加，枚举减

注意如果第i位加了a，第i+1位加了b，第i+2位加了c，那么a>=b>=c这个关系不能错

#include<stdio.h>#include<string.h>#define inf 1000000000;int dp[1005][10][10];int min(int x,int y){return x<y?x:y;}int main(){int i,j,k,l,a,c,b,len;char s1[1005],s2[1005];while(scanf("%s%s",s1,s2)!=EOF){len=strlen(s1);for(i=0;i<=len;i++)for(j=0;j<10;j++)for(k=0;k<10;k++)dp[i][j][k]=inf;dp[0][0][0]=0;for(i=0;i<len;i++)for(j=0;j<10;j++)for(k=0;k<10;k++){c=(20+s2[i]-s1[i]-j)%10;//向上for(a=0;a<=c;a++)for(b=0;b<=a;b++){dp[i+1][(k+a)%10][b]=min(dp[i+1][(k+a)%10][b],dp[i][j][k]+c);//最多需要c步。//当前状态的k到了i+1状态时到了对应的j位置}c=(10-c)%10;//向下for(a=0;a<=c;a++)for(b=0;b<=a;b++){dp[i+1][(k-a+10)%10][(10-b)%10]=min(dp[i+1][(k-a+10)%10][(10-b)%10],dp[i][j][k]+c);}}printf("%d\n",dp[len][0][0]);}return 0;}