ural 1145. Rope in the Labyrinth

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

 

Tags: graph theory  (hide tags for unsolved problems)

Difficulty: 425

 

题意：.是可以走的，#不可以走。所有的.组成了一棵树，问这棵树的直径。

分析：求树的直径。bfs即可。

 

 /**
 Create By yzx - stupidboy
 */
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 #include <iomanip>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair

 inline int Getint()
 {
     int Ret = ;
     char Ch = ' ';
     bool Flag = ;
     while(!(Ch >= '' && Ch <= ''))
     {
         if(Ch == '-') Flag ^= ;
         Ch = getchar();
     }
     while(Ch >= '' && Ch <= '')
     {
         Ret = Ret *  + Ch - '';
         Ch = getchar();
     }
     return Flag ? -Ret : Ret;
 }

 const int N = ;
 const int DX[] = {-, , , }, DY[] = {, -, , };
 int n, m;
 char graph[N][N];
 int dp[N][N];
 queue<pair<int, int> > que;

 inline void Input()
 {
     scanf("%d%d", &m, &n);
     for(int i = ; i < n; i++) scanf("%s", graph[i]);
 }

 inline bool Check(int x, int y)
 {
     if(x <  || y <  || x >= n || y >= m) return ;
     if(graph[x][y] != '.') return ;
     return ;
 }

 inline void Bfs(int sx, int sy)
 {
     for(int i = ; i < n; i++)
         for(int j = ; j < m; j++)
             dp[i][j] = INF;
     que.push(mk(sx, sy));
     dp[sx][sy] = ;
     while(sz(que))
     {
         int ux = que.front().ft, uy = que.front().sd;
         que.pop();
         for(int t = ; t < ; t++)
         {
             int vx = ux + DX[t], vy = uy + DY[t];
             if(Check(vx, vy) && dp[vx][vy] > dp[ux][uy] + )
             {
                 dp[vx][vy] = dp[ux][uy] + ;
                 que.push(mk(vx, vy));
             }
         }
     }
 }

 inline void GetMax(int &px, int &py)
 {
     int mx = -INF;
     for(int i = ; i < n; i++)
         for(int j = ; j < m; j++)
             if(mx < dp[i][j] && dp[i][j] < INF)
             {
                 mx = dp[i][j];
                 px = i, py = j;
             }
 }

 inline void Solve()
 {
     bool flag = ;
     for(int i = ; i < n && !flag; i++)
         for(int j = ; j < m && !flag; j++)
             if(graph[i][j] == '.')
             {
                 Bfs(i, j);
                 flag = ;
             }

     int px, py;
     GetMax(px, py);
     Bfs(px, py);

     GetMax(px, py);
     printf("%d\n", dp[px][py]);
 }

 int main()
 {
     freopen("a.in", "r", stdin);
     Input();
     Solve();
     return ;
 }