poj 2251 BFS

标签：bfspan idtransmarksp   pspan idtransmarkspa   

Description

Input

Output

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

Source

给出一三维空间的地牢,要求求出由字符‘S‘到字符‘E‘的最短路径

移动方向可以是上，下，左，右，前，后，六个方向

用BFS求最短路 比DFS好;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>

using namespace std;

struct node
{
	int x,y,z;
}p,s,e,q;
queue<node>Q;
int n,m,h;
char a[100][100][100];  //三维
int vis[100][100][100];
int dx[]={1,-1,0,0,0,0};
int dy[]={0,0,1,-1,0,0};
int dz[]={0,0,0,0,-1,1};
int step[100][100][100];
int Judge(node q)
{
	if(q.x>=0&&q.y>=0&&q.z>=0&&q.x<n&&q.y<m&&q.z<h&&a[q.x][q.y][q.z]!='#') //存在E
		return 1;
	return 0;
}
int BFS()
{
	while(!Q.empty())   //  <span id="transmark"></span>队列必须清空
		Q.pop();
	
	Q.push(s);
	vis[s.x][s.y][s.z]=1;
	step[s.x][s.y][s.z]=0;
	while(!Q.empty())
	{
		p=Q.front();
		Q.pop();
		for(int i=0;i<6;i++)
		{
			q.x=dx[i]+p.x;
			q.y=dy[i]+p.y;
			q.z=dz[i]+p.z;
			if(!Judge(q))
				continue;
			if(vis[q.x][q.y][q.z]==0)
			{
				Q.push(q);
				step[q.x][q.y][q.z]=step[p.x][p.y][p.z]+1;
				vis[q.x][q.y][q.z]=1;
			}
			if(a[q.x][q.y][q.z]=='E')
				return step[q.x][q.y][q.z];
		}
	}
	return -1;
}
int main()
{
	while(~scanf("%d%d%d",&n,&m,&h))
	{
		int flag=1;
		if(!n&&!m&&!h)
		break;
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				scanf("%s",a[i][j]);
				if(flag)
				for(int k=0;k<h;k++)
				{
					if(a[i][j][k]=='S')
					{
						s.x=i;s.y=j;s.z=k;
						flag=0;
						break;
					}
				}
			}
		}
		int x=BFS();
		if(x==-1)
		{
			printf("Trapped!\n");
	        continue;
		}
		printf("Escaped in %d minute(s).\n",x);
	}
}

poj 2251 BFS

标签：bfspan idtransmarksp   pspan idtransmarkspa   

原文地址：http://blog.csdn.net/became_a_wolf/article/details/48000499