【BFS】【HDOJ-1195】Open the Lock

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

 

思路：
BFS  可以有+1 -1 交换三种线路 进行BFS

 

参考代码：

#include <iostream>
#include <string>
#include <queue>
using namespace std;
char s1[5],s2[5];
int a[4],b[4];
int bfs();
int mark[10000];
typedef struct
{
    int x[4];
    int step;
}node;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>s1>>s2;
        int i;
        for(i=0;i<4;i++)
        {
            a[i]=s1[i]-'0';
            b[i]=s2[i]-'0';
        }
        cout<<bfs()<<endl;
    }
    return 0;
}

int bfs()
{
    memset(mark,0,sizeof(mark));
    node temp;
    queue<node> q;
    temp.step=0;
    int i;
    for(i=0;i<4;i++)
        temp.x[i]=a[i];
    q.push(temp);
    mark[a[0]*1000+a[1]*100+a[2]*10+a[3]]=1;
    while(!q.empty())
    {
        for(i=0;i<11;i++)
        {
            if(i<4)//加
            {
                node newl=q.front();
                newl.x[i]=q.front().x[i]+1;
                if(newl.x[i]==10)
                    newl.x[i]=1;
                if(!mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]])
                {
                    mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]]=1;
                    newl.step=q.front().step+1;
                    q.push(newl);
                    if(newl.x[0]==b[0]&&newl.x[1]==b[1]&&newl.x[2]==b[2]&&newl.x[3]==b[3])
                        return newl.step;
                }
            }
            else if(i<8)//减
            {
                node newl=q.front();
                newl.x[i-4]=q.front().x[i-4]-1;
                if(newl.x[i-4]==0)
                    newl.x[i-4]=9;
                if(!mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]])
                {
                    mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]]=1;
                    newl.step=q.front().step+1;
                    q.push(newl);
                    if(newl.x[0]==b[0]&&newl.x[1]==b[1]&&newl.x[2]==b[2]&&newl.x[3]==b[3])
                        return newl.step;
                }
            }
            else
            {
                node newl=q.front();
                int temp;
                temp=q.front().x[i-7];
                newl.x[i-8]=temp;
                newl.x[i-7]=q.front().x[i-8];
                if(!mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]])
                {
                    mark[newl.x[0]*1000+newl.x[1]*100+newl.x[2]*10+newl.x[3]]=1;
                    newl.step=q.front().step+1;
                    q.push(newl);
                    if(newl.x[0]==b[0]&&newl.x[1]==b[1]&&newl.x[2]==b[2]&&newl.x[3]==b[3])
                        return newl.step;
                }
            }
        }
        q.pop();
    }
}