HDU 1195 Open the Lock 暴力过法 DHOJ 1195

题目链接如下

http://acm.hdu.edu.cn/showproblem.php?pid=1195

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

Author

YE, Kai

Source

Zhejiang University Local Contest 2005

Recommend

Ignatius.L

/*
考虑到此题密码变换规则相当复杂，但是数据并不是很大，密码只有4位，结果只要求输出最短的步数，可以考虑穷举法
假设结果是X种变换，我们第一步到第X步考虑每一种变换，每一种变换只有11种可能，变换后的密码又有11种可能，
如此重复，直到正确密码出现，然后直接输出变换的次数。

这里我们用 vis[10000] 这个数组表示每一种密码，
step[10000]数组保存变换到每一种密码所需要的最小步数
a[5],b[5]数组保存每个密码数字的值
q队列更新每次的密码值，且来控制循环
其实也就是bfs只是写法不同而已
*/
#include <iostream>
#include <queue>
using namespace std;

int n,m;
int step[10000];
bool vis[10000];

int main()
{
int N;
cin>>N;
while(N--)
{
cin>>n>>m;
if(n==m) cout<<"0"<<endl;
int i,flag,temp;
int a[5],b[5];
queue<int> q;
q.push(n);
memset(step,0,sizeof(step));
memset(vis,true,sizeof(vis));
while(!q.empty())
{
temp=q.front();
q.pop();
b[4]=a[4]=temp%10;
b[3]=a[3]=temp/10%10;
b[2]=a[2]=temp/100%10;
b[1]=a[1]=temp/1000;
for (i=0;i<11;i++)
{
switch (i)
{
case 0:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[1]==9) b[1]=1;
else b[1]+=1;
}
break;
case 1:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[2]==9) b[2]=1;
else b[2]+=1;
}
break;
case 2:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[3]==9) b[3]=1;
else b[3]+=1;
}
break;
case 3:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[4]==9) b[4]=1;
else b[4]+=1;
}
break;
case 4:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[1]==1) b[1]=9;
else b[1]-=1;
}
break;
case 5:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[2]==1) b[2]=9;
else b[2]-=1;
}
break;
case 6:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[3]==1) b[3]=9;
else b[3]-=1;
}
break;
case 7:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
if (a[4]==1) b[4]=9;
else b[4]-=1;
}
break;
case 8:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
flag=b[1];
b[1]=a[2];
b[2]=flag;
}
break;
case 9:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
flag=b[2];
b[2]=a[3]; b[3]=flag;
}
break;
case 10:
{
b[1]=a[1];b[2]=a[2];b[3]=a[3];b[4]=a[4];
flag=b[3];
b[3]=a[4]; b[4]=flag;
}
break;
}
n=b[1]*1000+b[2]*100+b[3]*10+b[4];
if (vis[n])
{
vis[n]=false;
step[n]=step[temp]+1;
q.push(n);
if (n==m) { cout<<step[n]<<endl; break;}
}
}
}
}
}