POJ 1383 Labyrinth （树的直径求两点间最大距离）

Description

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

 

输入时先判断str[][]是否等于'.'，若等于记录'.'的位置bx，by，对bx,by进行搜索，找出离bx，by距离最大的点kx，ky，然后对kx，ky进行搜索找出最大的距离。

从数组中任意一个点开始搜索找到离他最远的点（x,y)，则数组中离(x,y)最远的点与（x，y）的距离就是任意两点距离的最大值。

 #include<cstdio>
 #include<queue>
 #include<string.h>
 #define M 1010
 using namespace std;
 int n,m,i,ans,flag[M][M],j,bx,by,kx,ky;
 char str[M][M];
 int dx[]={-,,,};
 int dy[]={,,-,};
 struct stu
 {
     int x,y,step;
 }st;
 void bfs(int xx,int yy)
 {
     memset(flag,,sizeof(flag));
     stu next;
     st.x=xx;
     st.y=yy;
     st.step=;
     queue<stu>que;
     flag[xx][yy] = ;
     que.push(st);
     while(!que.empty())
     {
         st=que.front();
         que.pop();
         for(int i =  ; i < ; i++)
         {
             next.x=st.x+dx[i];
             next.y=st.y+dy[i];
             next.step=st.step+;
             if(str[next.x][next.y] !='#' && next.x>= && next.y>= && next.x<m&&next.y<n&& flag[next.x][next.y] == )
             {
                 flag[next.x][next.y]=;
                 if(ans < next.step)
                 {
                     ans=next.step;
                     kx=next.x;
                     ky=next.y;
                 }
                 que.push(next);
             }
         }
     }
 }
 int main()
 {
     int t,k;
     scanf("%d",&t);
     while(t--)
     {
         ans=;

         k=;
         scanf("%d %d",&n,&m);
         for(i =  ; i < m ; i++)
         {
             scanf("%s",str[i]);
             if(k) continue;
             for(j =  ; j < n ; j++)
             {
                 if(str[i][j] == '.')
                 {
                     bx=i;
                     by=j;
                     k=;
                 }
             }
         }
         bfs(bx,by);            //搜索出离bx，by距离最远的点kx，ky
         bfs(kx,ky);            //搜索出两点之间最大的距离
         printf("Maximum rope length is %d.\n",ans);
     }
  }