Open the Lock （BFS）-优快云博客

本文链接：https://blog.youkuaiyun.com/J____xueming/article/details/77102167

本文介绍了一个使用广度优先搜索(BFS)算法解决四数字密码锁问题的方法。目标是在最少步骤内从初始状态到达指定密码状态。通过增加或减少每个数字上的数值以及交换相邻数字来改变密码组合。

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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

BFS

AC代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<vector>
#include<cstdlib>
using namespace std;
typedef long long ll;
int a,b,step;
int vis[100010]={0};
queue<int> q;
int bfs(int a,int b){
    q.push(a);
    vis[a]=1;
    while(!q.empty()){
        int len=q.size();
        while(len--){
            int num[5];
            int fir=q.front(),tmp;
            q.pop();
            if(fir==b)return step;
            num[3]=fir%10;
            num[2]=fir/10%10;
            num[1]=fir/100%10;
            num[0]=fir/1000;
            for(int i=0;i<4;i++){
                int m=num[i];
                num[i]+=1;
                if(num[i]>9)num[i]=1;
                tmp=0;
                for(int j=0;j<4;j++)
                    tmp=tmp*10+num[j];
                if(!vis[tmp]){
                    vis[tmp]=1;
                    q.push(tmp);
                }
                num[i]=m;
                num[i]-=1;
                if(num[i]<1)num[i]=9;
                tmp=0;
                for(int j=0;j<4;j++)
                    tmp=tmp*10+num[j];
                if(!vis[tmp]){
                    vis[tmp]=1;
                    q.push(tmp);
                }
                num[i]=m;
            }
            tmp=num[1]*1000+num[0]*100+num[2]*10+num[3];
            if(!vis[tmp]){
                vis[tmp]=1;
                q.push(tmp);
            }
            tmp=num[0]*1000+num[2]*100+num[1]*10+num[3];
            if(!vis[tmp]){
                vis[tmp]=1;
                q.push(tmp);
            }
            tmp=num[0]*1000+num[1]*100+num[3]*10+num[2];
            if(!vis[tmp]){
                vis[tmp]=1;
                q.push(tmp);
            }
        }
        step++;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        while(!q.empty())
            q.pop();
        step=0;
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&a,&b);
        bfs(a,b);
        printf("%d\n",step);
    }
    return 0;
}