ural 1145 Rope in the Labyrinth 图中 bfs求树的直径

1145. Rope in the Labyrinth

Time limit: 0.5 second
Memory limit: 64 MB

A labyrinth with rectangular form and size  m ×  n is divided into square cells with sides' length 1 by lines that are parallel with the labyrinth's sides. Each cell of the grid is either occupied or free. It is possible to move from one free cell to another free cells that share a common side with the cell. One cannot move beyond the labyrinth's borders. The labyrinth is designed pretty specially: for any two cells there is only one way to move from one cell to the other. There is a hook at each cell's center. In the labyrinth there are two special free cells, such that if you can connect the hooks of those two cells with a rope, the labyrinth's secret door will be automatically opened. The problem is to prepare a shortest rope that can guarantee, you always can connect the hooks of those two cells with the prepared rope regardless their position in the labyrinth.

Input

The first line contains integers  n and  m (3 ≤  n,  m ≤ 820). The next lines describe the labyrinth. Each of the next  m lines contains  n characters. Each character is either "#" or ".", with "#" indicating an occupied cell, and "." indicating a free cell.

Output

Print out in the single line the length (measured in the number of cells) of the required rope.

Sample

input	output
7 6 ####### #.#.### #.#.### #.#.#.# #.....# #######	8

首先我们要知道怎么求树的直径。

树的直径是指树的最长简单路。

求法: 两遍BFS :先任选一个起点BFS找到最长路的终点，再从终点进行BFS，则第二次BFS找到的最长路即为树的直径；

题意：给了个图，' . ' 可以走 ‘ # ’ 是墙。因为题目中规定了两点之间最多只有一条路可以走，而且必有一条路。可以见‘ . ’ 是一个树的结构。要求得距离最长的两点的距离。也就是求树的直径了。

做法： 找到节点，节点就是三面有‘ # ‘ 的 点。然后bfs 两遍求出直径就可以了。 

注意：图很大，dfs 会超内存。

#include<stdio.h>
#include<iostream>
#include<string>
#include<map>
#include<queue>
using namespace std;

int dir[4][2]={1,0,-1,0,0,1,0,-1};
char mp[830][830]; 
int ans;
int n,m;
int bix;
int biy;
int pp;

struct nodes
{
	int x,y,num,ff;
};
nodes now,tem;
bool ok(int a,int b)
{
    if(a==0&&b==1)
        return 0;
    if(a==1&&b==0)
        return 0;
    if(a==2&&b==3)
        return 0;
    if(a==3&&b==2)
        return 0;
    return 1;
}
void bfs(int x,int y)
{
	queue<nodes>q;
	now.ff=-1;
	now.num=0;
	now.x=x;
	now.y=y;
	q.push(now);

	while(!q.empty())
	{
		now=q.front();
		q.pop();
		if(now.num>ans)
		{
			bix=now.x;
			biy=now.y;
			ans=now.num;
		}
		int xx,yy;
		for(int i=0;i<4;i++)
		{
			xx=now.x+dir[i][0];
			yy=now.y+dir[i][1];
			if(xx>=0&&xx<n&&yy>=0&&yy<m&&mp[xx][yy]=='.'&&ok(i,now.ff))
			{
				tem.x=xx;
				tem.y=yy;
				tem.num=now.num+1;
				tem.ff=i;
				q.push(tem);

			}
		} 
	} 
}
int main()
{

	while(scanf("%d%d",&m,&n)!=EOF)
	{
	    ans=0;
	    for(int i=0;i<n;i++)//mp[n][m]
            scanf("%s",mp[i]);

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(mp[i][j]=='.')
                {
                    int num=0;
                    for(int k=0;k<4;k++)
                    {
                        int xx=i+dir[k][0];
                        int yy=j+dir[k][1];
                        if(xx>=0&&xx<n&&yy>=0&&yy<m)
                        {
                            if(mp[xx][yy]=='.')
                                num++;
                        }
                    } 
					if(num==1)
						pp=i*m+j;
                }
			}
		} 
		bfs(pp/m,pp%m);  
		bfs(bix,biy);
        printf("%d\n",ans);
	}
	return 0;
}