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;
}
