hdu 1195 Open the Lock 暴力穷举

Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3972    Accepted Submission(s): 1728

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:  数字只有四个,穷举才24种.  用逆序数来计算移动了几次.  

#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;


int main()
{
	int t,i,j,k;
	int tem1,tem2;
	char a[20],b[20];//输入的数
	int n[20],m[20];//把输入的存成int
	int num[40];
	int nixu[10];// 0 1 2 3  开始 不断变化  求逆序数
	int sumnixu;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",a);
		scanf("%s",b);
		for(i=0;i<4;i++)
		{
			n[i]=a[i]-'0';
			m[i]=b[i]-'0';
		    nixu[i]=i;
		}
		memset(num,0,sizeof(num));
		int minop=999999;
		for(i=0;i<24;i++)//第几种排序
		{
			sumnixu=0;//这种排序下 逆袭数大小,也就是exchange了几次
			for(j=0;j<4;j++)
			{
				
				tem2=max(n[nixu[j]],m[j]);
				tem1=min(n[nixu[j]],m[j]);//nixu数组 存的0123  .变化后代表n的下标.同时可以用来求逆序数
				num[i]+=min(tem2-tem1,9-tem2+tem1);
			    for(k=j+1;k<4;k++)//计算逆序数;
				{
					if(nixu[k]<nixu[j])
						sumnixu++;
				}
				
			}
			next_permutation(nixu,nixu+4); //全排列函数  头文件是algorithm .
			//printf("%d\n",sumnixu);
			num[i]+=sumnixu;
		    if(minop>num[i])
				minop=num[i];
		}
		printf("%d\n",minop);
	}
	return 0;
}