UVA11624 Fire!

11624
Fire!
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

//题意是，J在迷宫里，同时迷宫里有若干火源（数量大于等于1）同时火会周围蔓延，问人能否逃离，逃离就是有路可以到达边缘

//类似于双起点的BFS，可以看成多起点的搜索，先处理一下火能到的点得时间，在从人开始搜索

#include<bits/stdc++.h>using namespace std;const int maxn=1015;const int inf=200000;#define lson rt<<1,l,m#define rson rt<<1|1,m+1,r#define For(i,n) for(int i=0;i<(n);i++)template<class T>inline T read(T&x){    char c;    while((c=getchar())<=32);    bool ok=false;    if(c=='-')ok=true,c=getchar();    for(x=0; c>32; c=getchar())        x=x*10+c-'0';    if(ok)x=-x;    return x;}template<class T> inline void read_(T&x,T&y){    read(x);    read(y);}template<class T> inline void write(T x){    if(x<0)putchar('-'),x=-x;    if(x<10)putchar(x+'0');    else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){    write(x);    putchar('\n');}//  -------IO template------struct node{    int x,y;    node(int a,int b)    {        x=a;        y=b;    }};int dx[]= {0,0,1,-1};int dy[]= {1,-1,0,0};int n,m;char a[maxn][maxn];int  d[maxn][maxn];vector<node> v;void bfs_F()//相当于多起点的搜索 BFS是一层一层的进行，正好达到了“同时”进行搜索的目标{    queue<node> q;    int kk=v.size();    For(i,kk)q.push(v[i]);    while(!q.empty())    {        node s=q.front();        q.pop();        for(int i=0; i<4; i++)        {            int xx=dx[i]+s.x;            int yy=dy[i]+s.y;            if(xx>n||xx<0||yy>m||yy<0)continue;            if(a[xx][yy]=='#')continue;            if(d[xx][yy])continue;            q.push(node(xx,yy));            d[xx][yy]=d[s.x][s.y]+1;        }    }}int dd[maxn][maxn];bool bfs_J(int x,int y){    queue<node> q;    q.push(node(x,y));    memset(dd,0,sizeof(dd));    dd[x][y]=1;    while(!q.empty())    {        node s=q.front();        q.pop();        if(s.x==0||s.x==n-1||s.y==0||s.y==m-1)        {            printf("%d\n",dd[s.x][s.y]);            return true;        }        for(int i=0; i<4; i++)        {            int xx=dx[i]+s.x;            int yy=dy[i]+s.y;            if(xx<0||yy<0||xx>n||yy>m)continue;            if(a[xx][yy]=='#')continue;            if(d[xx][yy]&&dd[s.x][s.y]+1>=d[xx][yy])continue;            if(dd[xx][yy])continue;            q.push(node(xx,yy));            dd[xx][yy]=dd[s.x][s.y]+1;            if(xx==0||xx==n-1||yy==0||yy==m-1)            {                printf("%d\n",dd[xx][yy]);                return true;            }        }    }    return false;}int main(){    int i,j,k,t;    int T;#ifndef ONLINE_JUDGE    freopen("test.txt","r",stdin);    freopen("Ab.txt","w",stdout);#endif // ONLINE_JUDGE    read(T);    while(T--)    {        read_(n,m);        v.clear();        int jx,jy,fx,fy;        for(int i=0; i<n; i++)        {            for(int j=0; j<m; j++)            {                d[i][j]=0;                scanf("%c",&a[i][j]);                if(a[i][j]=='J')                {                    jx=i;                    jy=j;                }                else if(a[i][j]=='F')                {                    v.push_back(node(i,j));                    d[i][j]=1;                }            }            getchar();        }        bfs_F();        if(!bfs_J(jx,jy))            printf("IMPOSSIBLE\n");    }    return 0;}