北京林业大学“计蒜客”杯程序设计竞赛 网络赛 马踏棋盘的问题

//使用bfs
#include<stdio.h>
#include<iostream>
#include<utility>
#include<queue>
#include<string.h>
using namespace std;

char a[100][100];
int d[100][100]; //存放从起点开始到达每个点所用步数 ,同时也可标记该点有没有走过

int n, m;
int dx[] = {-1,-2,-2, -1, 1, 2, 2, 1};
int dy[] = {-2, -1, 1, 2, 2, 1, -1, -2};
int sx, sy;
int rx, ry;
typedef pair<int , int> P;

int bfs();
int isOk(int x, int y);

int isOk(int x, int y)
{
if((x >= 0 && x < n) &&(y >= 0 && y < m) && d[x][y] == -1 && a[x][y] != '#')
return 1;
return 0;
}
int bfs()
{
queue<P> que;
int i;
que.push(P(sx, sy));
d[sx][sy] = 0;
//不断循环直到队列的长度为零
while(que.size())
{
int nx, ny;
P p = que.front(); que.pop();
//  printf("出栈（%d,%d)\n", p.first, p.second);
if(a[p.first][p.second] == 'e')
{

break;
}

for(i = 0; i < 8; i++)
{
nx = p.first + dx[i];
ny = p.second + dy[i];
//  printf("%d, %d\n", nx, ny);

if(isOk(nx, ny))
{
if((i == 1 || i == 2) && a[p.first-1][p.second] == '#')
continue;
else if((i == 3 || i == 4) && a[p.first][p.second+1] == '#')
continue;
else if((i == 5 || i == 6) && a[p.first+1][p.second] == '#')
continue;
else if((i == 7 || i == 0) && a[p.first][p.second-1] == '#')
continue;
que.push(P(nx, ny));
if(nx == 1 && ny == 0 )
{
printf("i = %d, %c\n", i, a[nx][ny-1]);
}
printf("入栈（%d,%d)\n", nx, ny);
d[nx][ny] = d[p.first][p.second]+1;
}
}
}
return d[rx][ry];
}
int main(void)
{
sx = 0;
sy = 0;
int i = 0, j =0 ;
while(scanf("%d%d", &n, &m) != EOF)
{
getchar();
for(i = 0; i < n; i++)
{

for(j = 0; j <m; j++)
{
a[i][j] = getchar();
}
getchar();
}
rx = n-1; ry = m-1;
memset(d, -1, sizeof(d));
d[sx][sy] = 0;
int res = bfs();
printf("%d\n", res);
}
return 0;

}