牛客练习赛12-B-迷宫

题目链接:https://www.nowcoder.net/acm/contest/
解题思路:最容易想到的肯定是BFS,这里主要考虑两种情况,
第一种是直接不需要拿到钥匙就可以到,第二种是需要拿到钥匙然后再去开锁,这里需要注意的是,
就算能直接到也需要判断那种方法用到的步数最少。有可能出错的地方是:1,把队列Q开在了全局,这时候需要每bfs一次就把Q清空。2,memset(v,0,sizeeof(v))可以写在bfs里面也可以每bfs一次就执行一次。3,因为迷宫我是从1开始的,所以最后没有边界的判断,如果是从0开始需要加上。

AC代码:
#include
#include
#include
#include
#include
#include
using namespace std;
int h, k;
char m[505][505];
int v[505][505];
int t[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };
int dx, dy, ex, ey, sx, sy, kx, ky;
int ans;
struct Node
{
	int i;
	int j;
	int stp;
};
Node now;
int bfs(int x, int y,int x2,int y2,int stp)
{
	memset(v, 0, sizeof(v));
	v[x][y] = 1;
	now.i = x;
	now.j = y;
	now.stp = stp;
	queue Q;
	Q.push(now);
	while (!Q.empty())
	{
		now = Q.front();
		Q.pop();
		Node next;
		if (now.i == x2&&now.j == y2)
		{
			return now.stp+1;
		}
		for (int i = 0; i < 4; i++)
		{
			int x = now.i + t[i][0];
			int y = now.j + t[i][1];
			if (m[x][y] == '.'&&v[x][y] == 0)
			{
				v[x][y] = 1;
				next.i = x;
				next.j = y;
				next.stp = now.stp + 1;
				Q.push(next);
			}
		}
	}
	return -1;
}
int main()
{
	scanf("%d%d", &h, &k);
	for (int i = 1; i <=h; i++)
	{
		scanf("%s", m[i]+1);
	}
	for (int i = 1; i <= h; i++)
	{
		for (int j = 1; j <= k; j++)
		{
			if (m[i][j] == 'S')
			{
				sx = i;
				sy = j;
			}
			if (m[i][j] == 'D')
			{
				dx = i;
				dy = j;
			}
			if (m[i][j] == 'K')
			{
				kx = i;
				ky = j;
			}
			if (m[i][j] == 'E')
			{
				ex = i;
				ey = j;
			}
		}
	}  
	m[sx][sy] = '.';
	m[ex][ey] = '.';
	m[kx][ky] = '.';
	int ans1 = bfs(sx, sy, ex, ey,0);
	m[kx][ky] = '.';
	int ans2 = bfs(sx, sy, kx, ky, 0);
	m[dx][dy] = '.';
	int ans3 = bfs(kx, ky, dx, dy, 0);
	int ans4 = bfs(dx, dy, ex, ey, 0);
	if (ans1 == -1)
	{
		if (ans2 == -1 || ans3 == -1 || ans4 == -1)
		{
			printf("-1\n");
		}
		else
		{
			printf("%d\n", ans2 + ans3 + ans4);
		}
	}
	else
	{
		ans = min(ans1, ans2 + ans3 + ans4);
		printf("%d\n", ans);
	}
	return 0;
}
有问题私聊哦~ 么么哒

你可能感兴趣的:(牛客竞赛)