HDU 1372 Knight Moves(最简单也是最经典的bfs)
传送门:
http://acm.hdu.edu.cn/showproblem.php?pid=1372
Knight Moves
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13384 Accepted Submission(s): 7831
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.
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
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.
#include<bits/stdc++.h>
using namespace std;
#define max_v 10
int G[max_v][max_v];
int dir[][]= {{,-},{,-},{,},{,},{-,},{-,},{-,-},{-,-}};
int step;
int sx,sy,fx,fy;
struct node
{
int x,y,step;
};
void bfs()
{
memset(G,,sizeof(G));
queue<node> q;
node p,next;
p.x=sx;
p.y=sy;
p.step=;
G[p.x][p.y]=;
q.push(p); while(!q.empty())
{
p=q.front();
q.pop(); if(p.x==fx&&p.y==fy)
{
step=p.step;
return ;
} for(int i=; i<; i++)
{
next.x=p.x+dir[i][];
next.y=p.y+dir[i][]; if(next.x>=&&next.y>=&&next.x<=&&next.y<=&&G[next.x][next.y]==)
{
next.step=p.step+;
G[next.x][next.y]=;
q.push(next);
}
}
}
}
int main()
{
char c1,c2;
int y1,y2;
while(~scanf("%c%d %c%d",&c1,&y1,&c2,&y2))
{
getchar();
sx=c1-'a'+;
sy=y1;
fx=c2-'a'+;
fy=y2;
bfs();
printf("To get from %c%d to %c%d takes %d knight moves.\n",c1,y1,c2,y2,step);
}
return ;
}
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