Joe works in a maze. Unfortunately, portions of the maze have caught on fire,
and the owner of the maze neglected to create a fire escape plan. Help Joe
escape the maze.
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 (not diagonally). The fire spreads all four directions from each square that is on fire. 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.
The first line of input contains a single integer, the number of test cases to follow. The first line of each test case contains the two integers R and C, separated by spaces, with 1 <= R,C <= 1000. The following R lines of the test case each contain one row of the maze. Each of these lines contains exactly C characters, and each of these characters is one of:
There will be exactly one J in each test case.
2 4 4 #### #JF# #..# #..# 3 3 ### #J. #.F
For each test case, output a single line containing IMPOSSIBLE if Joe cannot exit the maze before the fire reaches him, or an integer giving the earliest time Joe can safely exit the maze, in minutes.
3 IMPOSSIBLE
讲解：写了一下午还是没写出来，这种题还没有做过，的确很难想到用这种方法；
参考的别人的代码，两个搜索，比较人到这个地方时，火来了没有；
1 #include<stdio.h> 2 #include<string.h> 3 int T,m,n,ans,ok,step[1005][1005],bu[1005][1005]; // step用来初始化火，bu用来记录人 4 char mat[1005][1005]; 5 struct C 6 { 7 int x,y; 8 }q[10000100]; 9 int dx[4]={1,0,-1,0}; 10 int dy[4]={0,1,0,-1}; 11 void fire() 12 { 13 int fr=0,re=0; 14 memset(step,-1,sizeof(step)); //相当于标记访问，同时又记录步数 15 for(int i=0;i<n;i++) 16 for(int j=0;j<m;j++) 17 { 18 if(mat[i][j]==‘F‘) 19 { 20 q[re].x=i; 21 q[re++].y=j; 22 step[i][j]=0; 23 } 24 } 25 while(fr<re) { 26 int x=q[fr].x; 27 int y=q[fr].y; 28 for(int i=0;i<4;i++) 29 { 30 int nx=x+dx[i]; 31 int ny=y+dy[i]; 32 if(nx>=n || nx<0 || ny>=m || ny<0) continue; 33 if(step[nx][ny]!=-1) continue; 34 if(mat[nx][ny]==‘#‘) continue; 35 step[nx][ny]=step[x][y]+1; 36 q[re].x=nx; 37 q[re++].y=ny; 38 } 39 fr++; 40 } 41 } 42 void bfs() 43 { 44 memset(bu,-1,sizeof(bu)); //同理 45 int fr=0,re=0; 46 for(int i=0;i<n;i++) 47 for(int j=0;j<m;j++) 48 if(mat[i][j]==‘J‘) { 49 q[re].x=i; 50 q[re++].y=j; 51 bu[i][j]=0; 52 } 53 while(fr<re) 54 { 55 int nx=q[fr].x; 56 int ny=q[fr].y; 57 for(int i=0;i<4;i++) 58 { 59 int fx=nx+dx[i]; 60 int fy=ny+dy[i]; 61 if(nx==0 || nx==n-1 || ny==0 || ny==m-1) { 62 ok=1; 63 ans=bu[nx][ny]+1; 64 return ; 65 } 66 if(fx>=n || fx<0 || fy>=m || fy<0) continue; 67 if(bu[fx][fy]!=-1) continue; 68 if(mat[fx][fy]==‘#‘) continue; 69 70 /* 须特别注意，火有可能被墙包围，使得火无法蔓延。下句的意思是，如果火蔓延到这个格子， 71 并且火到这个格子的步数小于或等于人到这个格子的步数，那么跳过此格子。假使火根本没有蔓延过来 72 ，理所当然可以人走这个格子。 */ 73 if(step[fx][fy]!=-1 && bu[nx][ny]+1>=step[fx][fy]) continue; 74 q[re].x=fx; 75 q[re++].y=fy; 76 bu[fx][fy]=bu[nx][ny]+1; 77 78 } 79 fr++; 80 } 81 } 82 int main() 83 { 84 scanf("%d%*c",&T); 85 while(T--) { 86 scanf("%d %d%*c",&n,&m); 87 for(int i=0;i<n;i++) 88 gets(mat[i]); 89 fire(); 90 ok=0; 91 bfs(); 92 if(ok) printf("%d\n",ans); 93 else printf("IMPOSSIBLE\n"); 94 } 95 return 0; 96 }
UVA Problem B: Fire!,布布扣,bubuko.com
原文：http://www.cnblogs.com/lovychen/p/3689303.html