joj1185
1185: Knight Moves
Result TIME Limit MEMORY Limit Run Times AC Times JUDGE
1s 8192K 667 315 Standard
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 Specification
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 Specification
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.
这是一个典型的深度优先搜索的问题，用个move函数可以减少代码的重复，这是做前几个题的时候借鉴的几个师哥的
#include<cstdio>
#include<cstring>#include<queue>
using namespace std;
char str[10];
int visited[9][9];
class node
{
public:
node(int i=0,int j=0,int s=0):
x(i),y(j),step(s){}
int x,y;
int step;
};
int move[8][2]={{2,1},{2,-1},{1,2},{1,-2},
{-1,2},{-1,-2},{-2,1},{-2,-1}};
int bfs(node n1,node n2)
{
n1.step=0;
n2.step=0;
queue<node>q;
q.push(n1);
visited[n1.x][n1.y]=1;
while(!q.empty())
{
node temp;
n1=q.front();
q.pop();
//printf("%d %d %d\n",temp.x,temp.y,temp.step);
if(n1.x==n2.x&&n1.y==n2.y)
{
return n1.step;
}
int step=n1.step+1;
for(int i=0;i<8;i++)
{
int x=n1.x+move[i][0];
int y=n1.y+move[i][1];
if(x>=1&&x<=8&&y>=1&&y<=8&&!visited[x][y])
{
//printf("%d %d\n",x,y);
temp.x=x;temp.y=y;temp.step=step;
q.push(temp);
visited[x][y]=1;
// printf("%d\n",step);
}
}
}
return 0;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
while(gets(str))
{
memset(visited,0,sizeof(visited));
node n1,n2;
n1.x=str[0]-'a'+1;n1.y=str[1]-'0';
n2.x=str[3]-'a'+1;n2.y=str[4]-'0';
int step=bfs(n1,n2);
printf("To get from %c%c to %c%c takes %d knight moves.\n",str[0],str[1],str[3],str[4],step);
}
return 0;
}