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

inputoutput
7 6
#######
#.#.###
#.#.###
#.#.#.#
#.....#
#######

8
 
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。
 
  1 /**
  2 Create By yzx - stupidboy
  3 */
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 #include <cmath>
  8 #include <deque>
  9 #include <vector>
 10 #include <queue>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <map>
 14 #include <set>
 15 #include <ctime>
 16 #include <iomanip>
 17 using namespace std;
 18 typedef long long LL;
 19 typedef double DB;
 20 #define MIT (2147483647)
 21 #define INF (1000000001)
 22 #define MLL (1000000000000000001LL)
 23 #define sz(x) ((int) (x).size())
 24 #define clr(x, y) memset(x, y, sizeof(x))
 25 #define puf push_front
 26 #define pub push_back
 27 #define pof pop_front
 28 #define pob pop_back
 29 #define ft first
 30 #define sd second
 31 #define mk make_pair
 32 
 33 inline int Getint()
 34 {
 35     int Ret = 0;
 36     char Ch = ' ';
 37     bool Flag = 0;
 38     while(!(Ch >= '0' && Ch <= '9'))
 39     {
 40         if(Ch == '-') Flag ^= 1;
 41         Ch = getchar();
 42     }
 43     while(Ch >= '0' && Ch <= '9')
 44     {
 45         Ret = Ret * 10 + Ch - '0';
 46         Ch = getchar();
 47     }
 48     return Flag ? -Ret : Ret;
 49 }
 50 
 51 const int N = 1010;
 52 const int DX[] = {-1, 0, 1, 0}, DY[] = {0, -1, 0, 1};
 53 int n, m;
 54 char graph[N][N];
 55 int dp[N][N];
 56 queue<pair<int, int> > que;
 57 
 58 inline void Input()
 59 {
 60     scanf("%d%d", &m, &n);
 61     for(int i = 0; i < n; i++) scanf("%s", graph[i]);
 62 }
 63 
 64 inline bool Check(int x, int y)
 65 {
 66     if(x < 0 || y < 0 || x >= n || y >= m) return 0;
 67     if(graph[x][y] != '.') return 0;
 68     return 1;
 69 }
 70 
 71 inline void Bfs(int sx, int sy)
 72 {
 73     for(int i = 0; i < n; i++)
 74         for(int j = 0; j < m; j++)
 75             dp[i][j] = INF;
 76     que.push(mk(sx, sy));
 77     dp[sx][sy] = 0;
 78     while(sz(que))
 79     {
 80         int ux = que.front().ft, uy = que.front().sd;
 81         que.pop();
 82         for(int t = 0; t < 4; t++)
 83         {
 84             int vx = ux + DX[t], vy = uy + DY[t];
 85             if(Check(vx, vy) && dp[vx][vy] > dp[ux][uy] + 1)
 86             {
 87                 dp[vx][vy] = dp[ux][uy] + 1;
 88                 que.push(mk(vx, vy));
 89             }
 90         }
 91     }
 92 }
 93 
 94 inline void GetMax(int &px, int &py)
 95 {
 96     int mx = -INF;
 97     for(int i = 0; i < n; i++)
 98         for(int j = 0; j < m; j++)
 99             if(mx < dp[i][j] && dp[i][j] < INF)
100             {
101                 mx = dp[i][j];
102                 px = i, py = j;
103             }
104 }
105 
106 inline void Solve()
107 {
108     bool flag = 0;
109     for(int i = 0; i < n && !flag; i++)
110         for(int j = 0; j < m && !flag; j++)
111             if(graph[i][j] == '.')
112             {
113                 Bfs(i, j);
114                 flag = 1;
115             }
116 
117     int px, py;
118     GetMax(px, py);
119     Bfs(px, py);
120 
121     GetMax(px, py);
122     printf("%d\n", dp[px][py]);
123 }
124 
125 int main()
126 {
127     freopen("a.in", "r", stdin);
128     Input();
129     Solve();
130     return 0;
131 }
View Code

 

posted @ 2015-12-26 10:02  yanzx6  阅读(178)  评论(0)  编辑  收藏  举报