Open the Lock

C - Open the Lock

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 1195

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 
        
        
        
        

 

PS：题意为给两个四位数的数字N、M，

N的每一位都可以加一或减一或跟相邻的一位交换位置，每操作一次算一步

求N以上面操作方法之后变成M的最少步数

代码：

#include<iostream>
#include<stdio.h>
#include<queue>
#include<string>
#include<string.h>
using namespace std;

struct Node
{
	string str;
	int cou;
}node, temp;

string password;
int f[2] = { 1, -1 };
int dis[10][10][10][10] = { 0 };

queue<Node> nodes;

void bfs()
{
	while (!nodes.empty())
	{
		node = nodes.front();
		nodes.pop();
		if (node.str == password)
		{
			cout << node.cou << endl;
			while (!nodes.empty())
			{
				nodes.pop();
			}
			return ;
		}
		else
		{
			for (int i = 0; i < 4; ++i)
			{
				for (int j = 0; j < 2; ++j)
				{
					temp.str = node.str;
					int num = temp.str[i] - '0' + f[j];
					if (num == 0)
					{
						num = 9;
					}
					if (num == 10)
					{
						num = 1;
					}
					temp.str[i] = (num + '0');
					if (!dis[temp.str[0] - '0'][temp.str[1] - '0'][temp.str[2] - '0'][temp.str[3] - '0'])
					{
						dis[temp.str[0] - '0'][temp.str[1] - '0'][temp.str[2] - '0'][temp.str[3] - '0'] = 1;
						temp.cou = node.cou + 1;
						nodes.push(temp);
					}

					temp = node;
					int k = i + f[j];
					if (k > 0 && k < 4)
					{
						char c = temp.str[i];
						temp.str[i] = temp.str[k];
						temp.str[k] = c;
						if (!dis[temp.str[0] - '0'][temp.str[1] - '0'][temp.str[2] - '0'][temp.str[3] - '0'])
						{
							dis[temp.str[0] - '0'][temp.str[1] - '0'][temp.str[2] - '0'][temp.str[3] - '0'] = 1;
							temp.cou++;
							nodes.push(temp);
						}
					}
				}
			}
		}
	}
}

int main()
{
	int t;
	while (~scanf("%d", &t))
	{
		while (t--)
		{
			memset(dis, 0, sizeof(dis));
			cin >> node.str >> password;
			dis[node.str[0] - '0'][node.str[1] - '0'][node.str[2] - '0'][node.str[3] - '0'] = 1;
			node.cou = 0;
			nodes.push(node);
			bfs();
		}
	}

	return 0;
}