2015
e2 e4<br>a1 b2<br>b2 c3<br>a1 h8<br>a1 h7<br>h8 a1<br>b1 c3<br>f6 f6<br>
To get from e2 to e4 takes 2 knight moves.<br>To get from a1 to b2 takes 4 knight moves.<br>To get from b2 to c3 takes 2 knight moves.<br>To get from a1 to h8 takes 6 knight moves.<br>To get from a1 to h7 takes 5 knight moves.<br>To get from h8 to a1 takes 6 knight moves.<br>To get from b1 to c3 takes 1 knight moves.<br>To get from f6 to f6 takes 0 knight moves.<br>
思路:这道题其实挺好想,老师在课堂上讲过一个骑士救公主的题。这个比那个简单。但是我不会下国际象棋,所以对骑士怎么走刚开始不太清楚。国际象棋中,其实只能在一个范围内移动,对骑士的8个方向使用BFS算法即可;
AC代码:
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int step;
int to[8][2] = {-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2,-2,-1};//骑士移动的8个方位
int map[10][10],ex,ey;
char s1[5],s2[5];
struct node
{
int x,y,step;
};
int check(int x,int y)
{
if(x<0 || y<0 || x>=8 || y>=8 || map[x][y])
return 1;
return 0;
}
int bfs()
{
int i;
queue<node> Q;
node p,next,q;
p.x = s1[0]-'a';
p.y = s1[1]-'1';
p.step = 0;
ex = s2[0]-'a';
ey = s2[1]-'1';
memset(map,0,sizeof(map));
map[p.x][p.y] = 1;
Q.push(p);
while(!Q.empty())
{
q = Q.front();
Q.pop();
if(q.x == ex && q.y == ey)
return q.step;
for(i = 0;i<8;i++)
{
next.x = q.x+to[i][0];
next.y = q.y+to[i][1];
if(next.x == ex && next.y == ey)
return q.step+1;
if(check(next.x,next.y))
continue;
next.step = q.step+1;
map[next.x][next.y] = 1;
Q.push(next);
}
}
return 0;
}
int main()
{
while(~scanf("%s%s",s1,s2))
{
printf("To get from %s to %s takes %d knight moves.\n",s1,s2,bfs());
}
return 0;
}