POJ 2243解题报告

BFS

#include <stdio.h>
#include <memory.h>
#include <stdlib.h>
const int maxn=15;
int visit[maxn*maxn];
int dir[8][2]={{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2}};
char a[3];
char b[3];
int bfs(int st,int fin);
int main()
{
while(scanf("%s",a)!=EOF)
{

scanf("%s",b);
int sy=a[0]-'a';
int sx=a[1]-'1';
int fy=b[0]-'a';
int fx=b[1]-'1';
memset(visit,0,sizeof(visit));
int sl=sx*8+sy;
int fl=fx*8+fy;
if(sl==fl)
{
printf("To get from %s to %s takes 0 knight moves.\n",a,b);
}
else
{
printf("To get from %s to %s takes %d knight moves.\n",a,b,bfs(sl,fl));

}
}
return 0;
}
int bfs(int st,int fin)
{
int que[maxn*maxn];
int font=0;
int rear=0;
int distance[maxn*maxn];
distance[st]=0;
que[rear++]=st;
while(font<rear)
{
int on=que[font++];
int x=on/8;
int y=on%8;
int i;
for(i=0;i<8;i++)
{
int dx=x+dir[i][0];
int dy=y+dir[i][1];
int on2=dx*8+dy;
if(dx>=0&&dx<8&&dy>=0&&dy<8&&!visit[on2])
{
visit[on2]=1;
distance[on2]=distance[on]+1;
que[rear++]=on2;
if(on2==fin) return distance[on2];

}

}

}
}