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.
Input
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:
• #, 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.
Output
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.
Sample Input
2
4 4
#JF#
#…#
#…#
3 3
#J.
#.F
Sample Output
3
IMPOSSIBLE
題意：
一個迷宮。
'.‘爲空地，’#'爲牆壁，‘J’爲 Joe的起始位置，'F’爲火的起始位置（可能不止一處）。
joe和火都能從四個方向走，問走出迷宮最短時間
思路：
兩次BFS，目的分別求出Joe和火到達任意一點的時間，
（火的時間不再存放在和橫縱座標同一結構體中，而是在以更直觀的數組中（即下面代碼的fire數組），目的是爲了之後更方便的與Joe到達該點的時間作比較）
#include <stdio.h>
#include <string.h>
#define INF 0x3f3f3f
int n, m;
struct node
{
int x, y;
} str[1000005], p, q;
int t[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0} };
char a[1002][1002];
int fire[1002][1002];
int joe[1002][1002];
int judge(int x, int y)
{
if( x>=0&&x<n && y>=0&&y<m && a[x][y]!='#') return 1;
else return 0;
}
int judge_re(int x, int y)
{
if(x<0 || x>=n || y<0 || y>=m) return 1;
else return 0;
}
void bfs_fire()
{
int i, j, in, out;
in = out = 0;
for(i=0; i<n; i++)
{
for(j=0; j<m; j++)
{
if(a[i][j]=='F')
{
p.x = i;
p.y = j;
fire[i][j] = 0;
str[in++] = p;
}
}
}
while(in>out)
{
p = str[out++];
for(i=0; i<4; i++)
{
q = p;
if( judge(q.x+t[i][0], q.y+t[i][1])
&&fire[q.x][q.y]+1<fire[q.x+t[i][0]][q.y+t[i][1]] )
{
q.x += t[i][0];
q.y += t[i][1];
str[in++] = q;
fire[q.x][q.y] = fire[p.x][p.y]+1;
}
}
}
}
void bfs_joe(int sx, int sy)
{
int i, in, out;
in = out = 0;
p.x = sx;
p.y = sy;
joe[sx][sy] = 0;
str[in++] = p;
while(in>out)
{
p = str[out++];
for(i=0; i<4; i++)
{
q = p;
if(judge_re(q.x+t[i][0], q.y+t[i][1]))
{
printf("%d\n", joe[q.x][q.y]+1);
return;
}
if( judge(q.x+t[i][0], q.y+t[i][1])
&& joe[q.x][q.y]+1 < fire[q.x+t[i][0]][q.y+t[i][1]]
&& joe[q.x][q.y]+1 < joe[q.x+t[i][0]][q.y+t[i][1]])
{
q.x += t[i][0];
q.y += t[i][1];
str[in++] = q;
joe[q.x][q.y] = joe[p.x][p.y] + 1;
}
}
}
printf("IMPOSSIBLE\n");
}
int main()
{
int t, i, j, sx, sy;
scanf("%d", &t);
while(t--)
{
scanf("%d %d", &n, &m);
for(i=0; i<n; i++)
{
getchar();
for(j=0; j<m; j++)
{
scanf("%c", &a[i][j]);
if(a[i][j]=='J')
{
sx = i;
sy = j;
}
}
}
memset(fire, INF, sizeof(fire));
memset(joe, INF, sizeof(joe));
bfs_fire();
bfs_joe(sx, sy);
}
return 0;
}
題外話：
做這道題的時候犯了一個低級的錯誤，明明n,m已經定義成了全局變量，但在主函數裏又定義了一次，導致BFS的時候n，m值錯了，運行沒報錯，有一段時間才找出來。