简单bfs

跳马问题

Description

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.  
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.  

Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.  

 

Input

The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.  

 

Output

For each test case, print one line saying "To get from xx to yy takes n knight moves.".  

 

Sample Input

     
     
     
     

      
      
      
      e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.
     
     
     
     

代码：

#include<iostream>
#include<cstring>
#include <algorithm>
#include<queue>
#include<vector>
#define Maxn 110
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef pair<int, int> pii;
//char mp[Maxn][Maxn];
char a,b,c,d;
int vis[Maxn][Maxn];
int dx[]={-2,-1,1,2,2,1,-2,-1};
int dy[]={-1,-2,-2,-1,1,2,1,2};
int judge(int x,int y)
{
    return x>0&&y>0&&x<=8&&y<=8;
}
struct Item {
    int x, y, step;
    Item(int _x ,int _y, int _s) :
        x(_x), y(_y) , step(_s) {}
};
int bfs(int x,int y)
{
    int ans=0;
    vis[x][y]=1;
    queue<Item> p;
    p.push(Item(x, y,0));
    int k=10;
    while(1){
        Item it=p.front();p.pop();
        //cout<<it.step<<endl;
        for(int i=0;i<8;i++){
            int tx=it.x+dx[i],ty=it.y+dy[i];
            if(!judge(tx,ty)) continue;
            if(tx==c-'a'+1&&ty==d-'0') return it.step+1;
            if(!vis[tx][ty]) {p.push(Item(tx,ty,it.step+1));}
           // cout<<it.x+dx[i]<<' '<<it.y+dy[i]<<endl;
        }
    }
}
int main()
{
    while (cin>>a>>b>>c>>d){
        if(a==c&&b==d){
            cout<<"To get from "<<a<<b<<" to "<<c<<d<<" takes 0 knight moves."<<endl;
            continue;
        }
        memset(vis,0,sizeof vis);
        //cout<<a-'a'+1<<' '<<b-'0'<<endl;
        //cout<<c-'a'+1<<' '<<d-'0'<<endl;
        cout<<"To get from "<<a<<b<<" to "<<c<<d<<" takes "<<bfs(a-'a'+1,b-'0')<<" knight moves."<<endl;
    }
    return 0;
}

注：这题非常适合用bfs。