HDU - 1195:Open the Lock(BFS)

这是一个关于解决四位数密码锁的紧急任务。初始密码和目标密码给出,每次可以对任意数字加1或减1,或者与相邻数字交换。目标是最小步数解锁。输入包含测试用例数量及两个四位数,输出最小步骤数。示例显示了两种情况的解。解题思路是采用宽度优先搜索(BFS)策略。

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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

题目描述

给出2行数字,通过改变第一行的数字,使得与第二行数字相同,每个数字可以+1、-1、与相邻数字交换,都增加一次操作次数,求最少操作次数

思路:最少操作次数就会想到广搜,这也是很典型的广搜,需要注意的时,在每次试探时,都需要为next赋值为当前循环的now,这是广搜算法中比较重要的一点(保证所有情况都能试探到)

代码如下

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <queue>

using namespace std;

struct node
{
    int num[4],step;
}pre,last;
int book[11][11][11][11];

void bfs()
{
    int i;
    node now,next;
    queue<node> q;
    now = pre;
    now.step = 0;
    book[now.num[0]][now.num[1]][now.num[2]][now.num[3]] = 1;
    q.push(now);
    while(!q.empty())
    {
        now = q.front();
        q.pop();
        if(now.num[0]==last.num[0]&&now.num[1]==last.num[1]&&now.num[2]==last.num[2]&&now.num[3]==last.num[3])
        {
            printf("%d\n",now.step);
            return;
        }
        //+1操作
        for(i = 0;i < 4;i++)
        {
            next = now;	//每次都为next赋值为最初的值
            next.num[i]++;
            if(next.num[i] == 10)
                next.num[i] = 1;
            if(!book[next.num[0]][next.num[1]][next.num[2]][next.num[3]])
            {
                book[next.num[0]][next.num[1]][next.num[2]][next.num[3]] = 1;
                next.step++;
                q.push(next);
            }
        }
        //-1操作
        for(i = 0;i < 4;i++)
        {
            next = now;
            next.num[i]--;
            if(next.num[i] == 0)
                next.num[i] = 9;
            if(!book[next.num[0]][next.num[1]][next.num[2]][next.num[3]])
            {
                book[next.num[0]][next.num[1]][next.num[2]][next.num[3]] = 1;
                next.step++;
                q.push(next);
            }
        }
        //换位操作
        for(i = 0;i < 3;i++)
        {
            next = now;
            next.num[i] = now.num[i+1];
            next.num[i+1] = now.num[i];
            if(!book[next.num[0]][next.num[1]][next.num[2]][next.num[3]])
            {
                book[next.num[0]][next.num[1]][next.num[2]][next.num[3]] = 1;
                next.step++;
                q.push(next);
            }
        }
    }
}

int main()
{
    int i,t;
    char s1[10],s2[10];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",s1,s2);
        for(i = 0;i < 4;i++)
        {
            pre.num[i] = s1[i] - '0';
            last.num[i] = s2[i] - '0';
        }
        memset(book, 0, sizeof(book));
        bfs();
    }
    return 0;
}
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