Open the Lock[HDU1195]

最新推荐文章于 2025-08-09 15:58:26 发布

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最新推荐文章于 2025-08-09 15:58:26 发布

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文章标签： java

原文链接：http://www.cnblogs.com/dramstadt/p/3202643.html

本文介绍了如何在有限步数内解开一个四位数的密码锁，通过加减操作和相邻数字互换，最终找到正确的密码。

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Open the Lock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2914 Accepted Submission(s): 1271

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

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#include<stdio.h>
#include<string.h>
#include<queue>
#include<iostream>
using namespace std;
class node
{
    public:
        int x[4];
        int time;
        int val()
        {
            return x[0]*1000+x[1]*100+x[2]*10+x[3];
        }
};
int S,P;
bool s[10000];
int bfs()
{
    memset(s,false,sizeof(s));
    s[S]=true;
    queue<node> q;
    while (!q.empty()) q.pop();
    node tmp;
    tmp.x[0]=S/1000;
    tmp.x[1]=(S-tmp.x[0]*1000)/100;
    tmp.x[2]=(S-tmp.x[0]*1000-tmp.x[1]*100)/10;
    tmp.x[3]=S%10;
    tmp.time=0;
    q.push(tmp);
    while (!q.empty())
    {
        node x=q.front();
        q.pop();
        int state;
        if (x.val()==P) return x.time;
        int i;
        for (i=0;i<4;i++)
        {
            tmp=x;
            if (tmp.x[i]==9) tmp.x[i]=1;
            else tmp.x[i]++;
            tmp.time=x.time+1;
            state=tmp.val();
            if (!s[state])
            {
                s[state]=true;
                q.push(tmp);
            }
            tmp=x;
            if (tmp.x[i]==1) tmp.x[i]=9;
            else tmp.x[i]--;
            tmp.time=x.time+1;
            state=tmp.val();
            if (!s[state])
            {
                s[state]=true;
                q.push(tmp);
            }
        }
        for (i=0;i<3;i++)
        {
            tmp=x;
            int t=tmp.x[i];
            tmp.x[i]=tmp.x[i+1];
            tmp.x[i+1]=t;
            tmp.time=x.time+1;
            state=tmp.val();
            if (!s[state])
            {
                s[state]=true;
                q.push(tmp);
            }
        }
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&S,&P);
        printf("%d\n",bfs());
    }
    return 0;
}

转载于:https://www.cnblogs.com/dramstadt/p/3202643.html