暑假集训第四周周三赛F - Knight Moves 骑士的移动 STL

F - Knight Moves

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 1372

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.
      
      
      
      

分析：

又是STL ，用到了队列，其实就是一个模板，但是这个真的很重要，需要我去耐心掌握

用一个数组区存储它的八个方向，去一个一个判断它的运动情况，找到最优解

以后尽量先把它的坐标变化写出来，然后再去判断是否到达边界，可能会方便的多

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#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
int  s[8][2]= {-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};
int m[10][10],x1,y1;
char s1[5],s2[5];
struct  node
{
    int x,y,step;
};
int check(int x,int y)
{
    if(x<0||y<0||x>=8||y>=8||m[x][y])
        return 1;
    return 0;
}
int BFS()
{
    queue<node> q;
    node p,n,k;
    p.x=s1[0]-'a';
    p.y=s1[1]-'1';
    p.step=0;
    x1=s2[0]-'a';
    y1=s2[1]-'1';
    memset(m,0,sizeof(m));
    m[p.x][p.y]=1;
    q.push(p);
    while(!q.empty())
    {
        k=q.front();
        q.pop();
        if(k.x==x1&&k.y==y1)
            return k.step;
        for(int i=0; i<8; i++)
        {
            n.x=k.x+s[i][0];
            n.y=k.y+s[i][1];
            if(n.x==x1&&n.y==y1)
                return k.step+1;
            if(check(n.x,n.y))
                continue;
            n.step=k.step+1;
            m[n.x][n.y]=1;
            q.push(n);
        }
    }
    return 0;
}
int main()
{
    while(scanf("%s%s",s1,s2)!=EOF)
        printf("To get from %s to %s takes %d knight moves.\n",s1,s2,BFS());
    return 0;
}