Labyrinth

Labyrinth

The northern part of the Pyramid contains a very large and complicated
labyrinth. The labyrinth is divided into square blocks, each of them
either filled by rock, or free. There is also a little hook on the
floor in the center of every free block. The ACM have found that two
of the hooks must be connected by a rope that runs through the hooks
in every block on the path between the connected ones. When the rope
is fastened, a secret door opens. The problem is that we do not know
which hooks to connect. That means also that the neccessary length of
the rope is unknown. Your task is to determine the maximum length of
the rope we could need for a given labyrinth.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers C and R (3 <= C,R <= 1000) indicating the number of columns and rows. Then exactly R lines follow, each containing C characters. These characters specify the labyrinth. Each of them is either a hash mark (#) or a period (.). Hash marks represent rocks, periods are free blocks. It is possible to walk between neighbouring blocks only, where neighbouring blocks are blocks sharing a common side. We cannot walk diagonally and we cannot step out of the labyrinth.
The labyrinth is designed in such a way that there is exactly one path between any two free blocks. Consequently, if we find the proper hooks to connect, it is easy to find the right path connecting them.

Output
Your program must print exactly one line of output for each test case. The line must contain the sentence “Maximum rope length is X.” where Xis the length of the longest path between any two free blocks, measured in blocks.

Sample Input
2
3 3

#.#

7 6
#######
#.#.###
#.#.###
#.#.#.#
#…#
#######

Sample Output
Maximum rope length is 0.
Maximum rope length is 8.
Hint
Huge input, scanf is recommended.
If you use recursion, maybe stack overflow. and now C++/c 's stack size is larger than G++/gcc

题目大意：
在图中找到距离最远的两个点，输出它们之间的距离。
任意两个点之间是连通的。

解题思路：
需要两次bfs或dfs;先找到任意一个点，用bfs或dfs找到距离它最远的点并标记，然后以这个点为起点，用bfs或dfs找到距离它最远的点，输出二者之间的距离。

Code:

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define N 1006
using namespace std;
char a[N][N];
int d[4][2]={1,0,-1,0,0,-1,0,1};//移动方向 
bool vis[N][N];//标记是否访问过 
int n,m,ans;
struct node {
	int x,y;
	int step;//用步数计算二者之间的距离 
	friend bool operator > (node a,node b){//步数大的排前面，但没什么用 
		return a.step<b.step;
	}
};
node st,ag;
bool check(int x,int y){//判断是否符合条件 
	if(x>=0&&x<n&&y>=0&&y<m&&!vis[x][y]&&a[x][y]!='#'){
		return 1;
	}
	return 0;
}
void bfs(node st){//bfs模板 
	queue<node>q;         
	q.push(st);
	while(!q.empty()){
		node no;
		no=q.front();
		q.pop();
		ag=no;
		ans=max(ans,no.step);
		for(int i=0;i<4;i++){
			node nx;
			nx.x=no.x+d[i][0];
			nx.y=no.y+d[i][1];
			if(check(nx.x,nx.y)){
				vis[nx.x][nx.y]=1;
				nx.step=no.step+1;
				q.push(nx);
			}
		}
	}
}
int main(){
	int t;
	cin>>t;
	while(t--){
		memset(vis,0,sizeof(vis));//消除之前的标记 
		ans=0;
		scanf("%d%d",&m,&n);
		for(int i=0;i<n;i++) scanf("%s",a[i]);
		for(int i=0;i<n;i++){
			for(int j=0;j<m;j++){
				if(a[i][j]=='.'){
					st.x=i,st.y=j;
					vis[st.x][st.y]=1;
					st.step=0;
					bfs(st);// 第一次查找 
					goto end;//goto跳出多层循环，也可以用flag标记跳出循环 
				}
				
			}
		}
		end:
		memset(vis,0,sizeof(vis));
		ans=0;
		ag.step=0;//每次查找前都要清除之前的数据 
		bfs(ag);//第二次查找 
		printf("Maximum rope length is %d.
",ans);//注意输出格式 
	}
	
	return 0;
}