Description

Problem B: Fire!

Given Joe's location in the maze and which squares of the maze are on fire,you must determine whether Joe can exit the maze before the fire reaches him,and how fast he can do it.

Joe and the fire each move one square per minute, vertically or horizontally (notdiagonally). The fire spreads all four directions from each square that is onfire. Joe may exit the maze from any square that borders the edge of the maze.Neither Joe nor the fire may enter a square that is occupied by a wall.

Input Specification

#, a wall ., a passable square J, Joe's initial position in the maze, which is a passable square F, a square that is on fire There will be exactly one J in each test case.

Sample Input

2
4 4
####
#JF#
#..#
#..#
3 3
###
#J.
#.F

Output Specification

For each test case, output a single line containing IMPOSSIBLE if Joe cannot exit the maze beforethe fire reaches him, or an integer giving the earliest time Joe can safely exit themaze, in minutes.

Output for Sample Input

3
IMPOSSIBLE
















图论基础题。。




AC代码：

#include 
   
    
#include 
    
      #include 
     
       #include 
      
        #include 
       
         using namespace std; const int maxn = 1010; int n, m; char g[maxn][maxn]; //用于存储整个图 queue
        
          > q; //用于bfs的队列 int a[maxn][maxn]; //用于存储每个格子开始起火的时间 int move[][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}; //用于上下左右走 void bfs1() //第一次遍历，预处理每个格子起火的时间 { memset(a, -1, sizeof(a)); //初始化 while(!q.empty()) q.pop(); //清空队列 for(int i=0; i
         
           tmp = q.front(); q.pop(); int x = tmp.first, y = tmp.second; for(int i=0; i<4; i++) { int t1 = x + move[i][0], t2 = y + move[i][1]; if(a[t1][t2] != -1) continue; if(t1 < 0 || t2 < 0 || t1 >= n || t2 >= m) continue; if(g[t1][t2] == '#') continue; a[t1][t2] = a[x][y] + 1; q.push(make_pair(t1, t2)); } } } int b[maxn][maxn]; int bfs2() //第二次遍历 { memset(b, -1, sizeof(b)); while(!q.empty()) q.pop(); for(int i=0; i
          
            tmp = q.front(); q.pop(); int x = tmp.first, y = tmp.second; if(x == 0 || y == 0 || x == n - 1 || y == m - 1) return b[x][y] + 1; for(int i=0; i<4; i++) { int t1 = x + move[i][0], t2 = y + move[i][1]; if(t1<0 || t2<0 || t1>=n || t2>=m)continue; if(b[t1][t2]!=-1)continue; if(g[t1][t2]=='#')continue; if(a[t1][t2] != -1 && b[x][y] + 1 >= a[t1][t2]) continue; b[t1][t2] = b[x][y] + 1; q.push(make_pair(t1, t2)); } } return -1; //没有路可走出去 } int main() { int T; scanf("%d", &T); while(T--) { scanf("%d %d", &n, &m); for(int i=0; i