POJ2251 Dungeon Master(BFS)

题意：

要逃出地牢，地牢是三维的，有上下层之分

要点：

就是将原本二维的BFS转换为三维的就可以了，求最短距离，这种BFS我还是可以做的

15188937

Seasonal

2251

Accepted

304K

32MS

C++

1469B

2016-02-22 22:10:22

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
using namespace std;
char map[35][35][35];
bool visit[35][35][35];
int l, r, c;
int d[6][3] = { { 1,0,0 },{ -1,0,0 },{ 0,1,0 },{ 0,-1,0 },{ 0,0,1 },{ 0,0,-1 } };//三维的改变数组
struct node
{
	int x, y, z;
	int count;
};

bool judge(int x, int y, int z)//限定的条件判断
{
	if (x >= 0 && x < l&&y >= 0 && y < r&&z >= 0 && z < c&&visit[x][y][z]&&map[x][y][z]!='#')
		return true;
	return false;
}

int bfs(int x, int y, int z)
{
	queue<node> que;
	node u;
	u.x = x; u.y = y; u.z = z;
	u.count = 0;
	que.push(u);
	visit[x][y][z] = false;
	while (!que.empty())
	{
		node temp=que.front();
		if (map[temp.x][temp.y][temp.z] == 'E')
			return temp.count;
		que.pop();
		for (int i = 0; i < 6; i++)
		{
			node v;
			v.x = temp.x + d[i][0];
			v.y = temp.y + d[i][1];
			v.z = temp.z + d[i][2];
			if (judge(v.x,v.y,v.z))
			{
				visit[v.x][v.y][v.z] = false;
				v.count = temp.count+1;
				que.push(v);
			}
		}
	}
	return 0;//如果出不来就输出0
}

int main()
{
	int i, j, k,sx,sy,sz;
	while (~scanf("%d%d%d", &l, &r, &c),l+r+c)
	{
		memset(map, 0, sizeof(map));
		memset(visit, true, sizeof(visit));
		for (i = 0; i < l; i++)
			for (j = 0; j < r; j++)
			{
				scanf("%s", map[i][j]);
				for (k = 0; k < c; k++)
				{
					if (map[i][j][k] == 'S') { sx = i; sy = j; sz = k; }
				}
			}
		int step=bfs(sx, sy, sz);
		if (step)
			printf("Escaped in %d minute(s).\n", step);
		else
			printf("Trapped!\n");
	}
	return 0;
}