hdu 1372 Knight Moves

最新推荐文章于 2019-08-18 16:29:15 发布

转载最新推荐文章于 2019-08-18 16:29:15 发布 · 57 阅读

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原文链接：http://www.cnblogs.com/nefu929831238/p/5537471.html

本文介绍了一种使用宽度优先搜索算法解决国际象棋棋盘上骑士从一个位置移动到另一个位置所需的最少步数问题的方法。通过具体实例展示了算法的实现过程。

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Problem 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.

//简单宽搜，这道是自己独立完成的~~(●'◡'●)

#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;
char a[2],b[2];
int s1,s2,e1,e2;
int visit[10][10];
int to[8][2]={{2,1},{-2,1},{-2,-1},{2,-1},{1,2},{1,-2},{-1,-2},{-1,2}};

struct node
{
    int x,y;
    int step;
};

int bfs()
{
  queue<node> q;
  node st,ed;
  st.x=s1;
  st.y=s2;
  st.step=0;
  q.push(st);
  memset(visit,0,sizeof(visit));
  visit[s1][s2]=1;
  while(!q.empty())
  {
      st=q.front();
      q.pop();
      if(st.x==e1&&st.y==e2)
      {
          printf("To get from %c%c to %c%c takes ",a[0],a[1],b[0],b[1]);
          cout<<st.step<<" knight moves."<<endl;
          return st.step;
      }
      for(int i=0;i<8;i++)
      {
          ed.x=st.x+to[i][0];
          ed.y=st.y+to[i][1];
          if(1<=ed.x&&ed.x<=8&&1<=ed.y&&ed.y<=8&&!visit[ed.x][ed.y])
          {
              visit[ed.x][ed.y]=1;
              ed.step=st.step+1;
              q.push(ed);
          }
      }
  }
}

int main()
{
    int ans;
    while(cin>>a>>b)
    {
        s2=a[0]-'a'+1;
        s1=a[1]-'0';
        e2=b[0]-'a'+1;
        e1=b[1]-'0';
        bfs();
       // ans=bfs();
        //cout<<ans<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/nefu929831238/p/5537471.html