http://hncu.acmclub.com/index.php?app=problem_title&id=111&problem_id=1101
题目描述
小明很喜欢下国际象棋,一天,他拿着国际象棋中的“马”时突然想到一个问题:
给定两个棋盘上的方格a和b,马从a跳到b最少需要多少步?
现请你编程解决这个问题。
提示:国际象棋棋盘为8格*8格,马的走子规则为,每步棋先横走或直走一格,然后再往外斜走一格。
输入格式
输入包含多组测试数据。每组输入由两个方格组成,每个方格包含一个小写字母(a~h),表示棋盘的列号,和一个整数(1~8),表示棋盘的行号。
输出
对于每组输入,输出一行“To get from xx to yy takes n knight moves.”。
样例输入
e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
样例输出
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.
思路:
简单的BFS,只要弄清楚马的走法就可以了,一开始字符数组开小了,蛋疼,以后果断要大方点啊
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std; struct node
{
int x,y,step;
}; int vis[8][8];
int sx,sy,ex,ey,ans;
int to[8][2]={-1,-2,-2,-1,-2,1,-1,2,1,2,2,1,2,-1,1,-2}; int check(int x,int y)
{
if(x<0 || y<0 || x>=8 || y>=8)
return 1;
if(vis[x][y])
return 1;
return 0;
} void bfs()
{
int i;
queue<node> Q;
node a,next;
a.x = sx;
a.y = sy;
a.step = 0;
vis[sx][sy] = 1;
Q.push(a);
while(!Q.empty())
{
a = Q.front();
Q.pop();
if(a.x == ex && a.y == ey)
{
ans = a.step;
return ;
}
for(i = 0;i<8;i++)
{
next = a;
next.x+=to[i][0];
next.y+=to[i][1];
if(check(next.x,next.y))
continue;
next.step = a.step+1;
vis[next.x][next.y] = 1;
Q.push(next);
}
}
return ;
} int main()
{
char ch1[10],ch2[10];
while(~scanf("%s%s",ch1,ch2))
{
sx = ch1[0]-'a';
sy = ch1[1]-'1';
ex = ch2[0]-'a';
ey = ch2[1]-'1';
memset(vis,0,sizeof(vis));
bfs();
printf("To get from %s to %s takes %d knight moves.\n",ch1,ch2,ans);
} return 0;
}