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
题意：有多个起火点，每个起火点可以往他四周烧，Joe在起火区域内，问Joe能不能从火场逃生。
分析：两次bfs就可解决，第一个bfs计算烧到当前格需要多少时间，第二个bfs计算Joe能否逃生
PS：注意是有多个起火点。。。
#include<bits/stdc++.h>
using namespace std;
const int N = 1010;
const int inf = 0x3f3f3f3f;
struct node{
int x, y, s;
};
int n, m,fx, fy, jx, jy, ans;
char s[N][N];
bool vis[N][N];
int e1[N][N], e2[N][N];
int Next[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
void bfs1(){
memset(vis, false, sizeof(vis));
queue<node> Q;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(s[i][j] == 'F'){
vis[i][j] = true;
Q.push((node){i, j, 0});
}
}
}
while(!Q.empty()){
int xx = Q.front().x;
int yy = Q.front().y;
int ss = Q.front().s;
Q.pop();
for(int k = 0; k <= 3; k++){
int tx = xx + Next[k][0];
int ty = yy + Next[k][1];
if(tx < 0 || ty < 0 || tx >= n || ty >= m || vis[tx][ty])
continue;
if(s[tx][ty] == '.'){
vis[tx][ty] = true;
Q.push((node){tx, ty, ss + 1});
e1[tx][ty] = ss + 1;
}
}
}
}
inline int bfs2(int x, int y){
memset(vis, false, sizeof(vis));
vis[x][y] = true;
e2[x][y] = 0;
queue<node> Q;
Q.push((node){x, y, 0});
while(!Q.empty()){
int xx = Q.front().x;
int yy = Q.front().y;
int ss = Q.front().s;
Q.pop();
if(xx == 0 || yy == 0 || xx == n - 1 || yy == m - 1){
return ss + 1;
}
for(int k = 0; k <= 3; k++){
int tx = xx + Next[k][0];
int ty = yy + Next[k][1];
if(tx < 0 || ty < 0 || tx >= n || ty >= m || vis[tx][ty])
continue;
if(s[tx][ty] == '.' && ss + 1 < e1[tx][ty]){
vis[tx][ty] = true;
Q.push((node){tx, ty, ss + 1});
e2[tx][ty] = ss + 1;
}
}
}
return -1;
}
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("in.txt", "r", stdin);
#endif // ONLINE_JUDGE
int t;
scanf("%d", &t);
while(t--){
memset(e1, inf, sizeof(e1));
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++){
scanf("%s", s[i]);
}
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(s[i][j] == 'J')
jx = i, jy = j;
}
}
bfs1();
int ans = bfs2(jx, jy);
if(ans == -1)
printf("IMPOSSIBLE
");
else
printf("%d
", ans);
}
return 0;
}