杭電1195 open the lock(検索問題)BFS

Open the Lock
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5504    Accepted Submission(s): 2456
Problem Description
Now an emergent task for you is to open a password lock. The password is consisted of four digits. Each digit is numbered from 1 to 9. 
Each time, you can add or minus 1 to any digit. When add 1 to '9', the digit will change to be '1' and when minus 1 to '1', the digit will change to be '9'. You can also exchange the digit with its neighbor. Each action will take one step.
Now your task is to use minimal steps to open the lock.
Note: The leftmost digit is not the neighbor of the rightmost digit.
 
Input
The input file begins with an integer T, indicating the number of test cases. 
Each test case begins with a four digit N, indicating the initial state of the password lock. Then followed a line with anotther four dight M, indicating the password which can open the lock. There is one blank line after each test case.
 
Output
For each test case, print the minimal steps in one line.
 
Sample Input

   
   
   
   

    
    
    
    2
1234
2144

1111
9999
   
   
   
   

 
Sample Output

   
   
   
   

    
    
    
    2
4
   
   
   
   
   
   
   
   


   
   
   
   

テーマ:
上の列を最小限の操作で下の列に変える.
操作は全部で3種類に分けられ、11種類の操作があります.

 zifuchuan  change(zifuchuan a,int b)
{
    a.step++;
    if (b<4)       // +
    {
        if(a.num[b]==9)a.num[b]=1;
        else a.num[b]++;
    }
    else if (b < 8)  // -
    {
        if (a.num[b%4]==1)a.num[b%4]=9;
        else a.num[b%4]=a.num[b%4]-1;
    }
    else    // swap
    {
        int tmp;
        b%=4;
        tmp=a.num[b];
        a.num[b]=a.num[b+1];
        a.num[b+1]=tmp;
    }
    return a;
}

ここのvis配列は4次元で使用されます.

struct zifuchuan
{
    int num[4];//    
    int step;
}now,nex;
int vis[10][10][10][10];//         .

そして完全なACコードです.

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
struct zifuchuan
{
    int num[4];
    int step;
}now,nex;
int vis[10][10][10][10];
int mubiao[4];
char s1[5];
char s2[5];
 zifuchuan  change(zifuchuan a,int b)
{
    a.step++;
    if (b<4)       // +
    {
        if(a.num[b]==9)a.num[b]=1;
        else a.num[b]++;
    }
    else if (b < 8)  // -
    {
        if (a.num[b%4]==1)a.num[b%4]=9;
        else a.num[b%4]=a.num[b%4]-1;
    }
    else    //  
    {
        int tmp;
        b%=4;
        tmp=a.num[b];
        a.num[b]=a.num[b+1];
        a.num[b+1]=tmp;
    }
    return a;
}
void bfs()
{
    queue<zifuchuan>s;
    memset(mubiao,0,sizeof(mubiao));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<4;i++)
    {
        now.num[i]=s1[i]-'0';
        mubiao[i]=s2[i]-'0';
    }
    now.step=0;
    vis[now.num[0]][now.num[1]][now.num[2]][now.num[3]]=1;
    s.push(now);
    while(!s.empty())
    {
        now=s.front();
        if(now.num[0]==mubiao[0]&&now.num[1]==mubiao[1]&&now.num[2]==mubiao[2]&&now.num[3]==mubiao[3])
        {
            printf("%d
",now.step);
            return  ;
        }
        s.pop();
        for(int i=0;i<11;i++)
        {
            nex=change(now,i);
            if(vis[nex.num[0]][nex.num[1]][nex.num[2]][nex.num[3]]==0)
            {
                s.push(nex);
                vis[nex.num[0]][nex.num[1]][nex.num[2]][nex.num[3]]=1;
            }
        }
    }
    return ;
    
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",s1,s2);
        bfs();
    }
}