[最短路]Open the Lock

Posted 2021-02-12 lllxq

题目描述

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.

 

输入

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.

 

输出

For each test case, print the minimal steps in one line.

 

样例输入

复制样例数据

2
1234
2144

1111
9999

样例输出

2
4

 bfs+优先队列求最短路

#include <iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;

int st,ed;
int vis[10005];
int tmp[5];
struct Node{
  int val,step;
  bool operator <(const Node b)const{
    return step>b.step;
  }
};
priority_queue<Node> q;

int bfs(){
  while(!q.empty()) q.pop();
  vis[st]=1;q.push(Node{st,0});
  while(!q.empty()){
    Node top=q.top();q.pop();
    int val=top.val,step=top.step;
    if(val==ed) return step;
    tmp[1]=(val/1000)%10,tmp[2]=(val/100)%10;
    tmp[3]=(val/10)%10,tmp[4]=(val/1)%10;
    int cpy;
    for(int i=1;i<=4;i++){
        cpy=tmp[i];
        if(tmp[i]==9) tmp[i]=1;
        else tmp[i]++;
        int now=tmp[1]*1000+tmp[2]*100+tmp[3]*10+tmp[4];
        if(!vis[now]) vis[now]=1,q.push(Node{now,step+1});
        tmp[i]=cpy;
    }
    for(int i=1;i<=4;i++){
        cpy=tmp[i];
        if(tmp[i]==1) tmp[i]=9;
        else tmp[i]--;
        int now=tmp[1]*1000+tmp[2]*100+tmp[3]*10+tmp[4];
        if(!vis[now]) vis[now]=1,q.push(Node{now,step+1});
        tmp[i]=cpy;
    }
    for(int i=1;i<=3;i++){
        swap(tmp[i],tmp[i+1]);
        int now=tmp[1]*1000+tmp[2]*100+tmp[3]*10+tmp[4];
        if(!vis[now]) vis[now]=1,q.push(Node{now,step+1});
        swap(tmp[i],tmp[i+1]);
    }
  }
  return -1;
}

int main()
{
    int t;scanf("%d",&t);
    while(t--){
        memset(vis,0,sizeof(vis));
        scanf("%d%d",&st,&ed);
        int ans=bfs();
        printf("%d
",ans);
    }
    return 0;
}

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