hdu 1195(搜索)

Posted 2020-07-23 AC菜鸟机

Open the Lock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5970    Accepted Submission(s): 2666

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

 

这种类型做的比较少，还是挺有收获的。

对每一位进行标记，然后对每种操作进行处理。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iostream>
using namespace std;
typedef long long LL;
bool vis[10][10][10][10];
struct Node{
    int v[4];
    int step;
};
Node s,t;
bool _equal(Node a,Node b){
    for(int i=0;i<4;i++){
        if(a.v[i]!=b.v[i]) return false;
    }
    return true;
}
Node operate(int x,Node now){
    Node next;
    for(int i=0;i<4;i++){
        next.v[i] = now.v[i];
    }
    next.step=now.step+1;
    if(x<4){ ///+
        if(now.v[x]==9) next.v[x]=1;
        else next.v[x]=now.v[x]+1;
    }else if(x<8){ ///-
        if(now.v[x%4]==1) next.v[x%4]=9;
        else next.v[x%4]=now.v[x%4]-1;
    }else{ ///exchange
        int a = now.v[x%4];
        int b = now.v[x%4+1];
        next.v[x%4]=b;
        next.v[x%4+1] = a;
    }
    return next;
}
int bfs(Node s){
    memset(vis,false,sizeof(vis));
    queue <Node>q;
    vis[s.v[0]][s.v[1]][s.v[2]][s.v[3]]=true;
    q.push(s);
    s.step = 0;
    while(!q.empty()){
        Node now = q.front();
        q.pop();
        if(_equal(now,t)){
            return now.step;
        }
        for(int i=0;i<11;i++){ ///总共11种操作,[1-4]+ [1-4]- exwchange[1,2][2,3][3,4]
            Node next=operate(i,now);
            if(vis[next.v[0]][next.v[1]][next.v[2]][next.v[3]]==false){
                vis[next.v[0]][next.v[1]][next.v[2]][next.v[3]]=true;
                q.push(next);
            }
        }
    }
    return -1;
}
int main()
{
    char s1[5],s2[5];
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){

        scanf("%s",s1);
        scanf("%s",s2);
        for(int i=0;i<4;i++){
            s.v[i]=s1[i]-‘0‘;
            t.v[i]=s2[i]-‘0‘;
        }
        int res = bfs(s);
        printf("%d\n",res);
    }

    return 0;
}