[宽度优先搜索] HDU 1372 Knight Moves

Knight Moves

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10542    Accepted Submission(s): 6211

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.

Source

University of
Ulm Local Contest 1996

原文大意：对一个国际象棋的棋盘，每次输入骑士（骑士=马）的起始位置和目标位置，问至少要多少步才能走到。

解题思路：其实是一个裸的广搜，贴上充数用的，输入输出打麻烦了。

#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

struct mp

  {

       int x,y,step;

       mp(int x,int y,int step):x(x),y(y),step(step){}

  };

queue<mp> q;

void bfs(int xbegin,int ybegin,int xend,int yend)

  {

       const int move[2][8]={{1,2,2,1,-1,-2,-2,-1},{2,1,-1,-2,-2,-1,1,2}};

       int i,x,y,visit[9][9];

       char sans1,sans2;

       memset(visit,0,sizeof(visit));

       if(xbegin==xend&&ybegin==yend)

         {

             sans1=xbegin+'a'-1;

             sans2=xend+'a'-1;

             printf("To get from %c%d to %c%d takes %d knight moves.\n",sans1,ybegin,sans2,yend,0);

             return;

          }

       while(!q.empty()) q.pop();

       q.push(mp(xbegin,ybegin,0));

       visit[xbegin][ybegin]=1;

       while(!q.empty())

         {

             for(i=0;i<8;++i)

               {

                x=q.front().x+move[0][i];y=q.front().y+move[1][i];

                if(x==xend&&y==yend)

                   {

                     sans1=xbegin+'a'-1;

                     sans2=xend+'a'-1;

                     printf("To get from %c%d to %c%d takes %d knight moves.\n",sans1,ybegin,sans2,yend,q.front().step+1);

                     return;

                   }

                if(x>0&&x<9&&y>0&&y<9&&(!visit[x][y]))

                 {

                             q.push(mp(x,y,q.front().step+1));

                             visit[x][y]=1;

                       }

                     }

             q.pop();

          }

       return;

  }

int main()

  {

       char s[5];

        int x_start,x_end,y_start,y_end;

       while(scanf("%s",&s)!=EOF)

         {

             x_start=s[0]-'a'+1;

             y_start=s[1]-'0';

             scanf("%s",s);

             x_end=s[0]-'a'+1;

             y_end=s[1]-'0';

             bfs(x_start,y_start,x_end,y_end);

          }

       return 0;

  }