HDU1195Open the Lock （暴力BFS广搜）

最新推荐文章于 2025-09-09 08:37:13 发布

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最新推荐文章于 2025-09-09 08:37:13 发布

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文章标签： java

原文链接：http://www.cnblogs.com/dshn/p/4750396.html

本文介绍了一种通过广度优先搜索(BFS)解决密码锁问题的方法。该问题要求从初始状态通过最少步骤达到目标状态，操作包括数字加减1及相邻数字交换。文章提供了完整的C++代码实现，并详细解释了如何使用四维数组记录状态。

Open the Lock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3847 Accepted Submission(s): 1661

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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题意：将第一行的4个数字变成第二行的数字
思路：利用广搜穷举所有情况。注意每一位可+1可-1,相邻两位可以交换位置,用一个visit[][][][]四维数组记录可能出现的数字串出现过没有。例如，"2143"出现过了，则 isv[2][1][4][3]=1。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
int a[5],b[5];
int visit[10][10][10][10];
struct nodes
{
    int num[5];
    int m;
};
nodes node1,node2;
int bfs()
{
    queue<nodes> q;
    while(!q.empty())
        q.pop();
    node1.num[0]=a[0];
    node1.num[1]=a[1];
    node1.num[2]=a[2];
    node1.num[3]=a[3];
    node1.m=0;
    visit[a[0]][a[1]][a[2]][a[3]]=1;
    q.push(node1);
    while(!q.empty())
    {
        node1=q.front();
        q.pop();
        if(node1.num[0]==b[0]&&node1.num[1]==b[1]&&node1.num[2]==b[2]&&node1.num[3]==b[3])
        return node1.m;
        for(int i=0;i<4;i++)
        {
            node2=node1;
            if(node1.num[i]==9)
            node2.num[i]=1;
            else
            node2.num[i]=node1.num[i]+1;
            if(!visit[node2.num[0]][node2.num[1]][node2.num[2]][node2.num[3]])
            {

                node2.m=node1.m+1;
                visit[node2.num[0]][node2.num[1]][node2.num[2]][node2.num[3]]=1;
                q.push(node2);
            }
            node2=node1;
            if(node1.num[i]==1)
            node2.num[i]=9;
            else
            node2.num[i]=node1.num[i]-1;
            if(!visit[node2.num[0]][node2.num[1]][node2.num[2]][node2.num[3]])
            {

                node2.m=node1.m+1;
                visit[node2.num[0]][node2.num[1]][node2.num[2]][node2.num[3]]=1;
                q.push(node2);
            }
        }
        for(int i=0;i<3;i++)
        {
            node2=node1;
            node2.num[i]=node1.num[i+1];
            node2.num[i+1]=node1.num[i];
           if(!visit[node2.num[0]][node2.num[1]][node2.num[2]][node2.num[3]])
            {

                node2.m=node1.m+1;
                visit[node2.num[0]][node2.num[1]][node2.num[2]][node2.num[3]]=1;
                q.push(node2);
            }
        }


    }
return 0;
}
int main()
{
    int t,t1,t2;
    scanf("%d",&t);
    while(t--)
    {
        memset(visit,0,sizeof(visit));
        scanf("%d%d",&t1,&t2);
        int i=0,j=0;
       while(t1>0)
       {
           a[i++]=t1%10;
           t1=t1/10;
       }
       while(t2>0)
       {
           b[j++]=t2%10;
           t2=t2/10;
       }
        printf("%d\n",bfs());
    }
}