先对火BFS一次，求出每个点的最小着火时间。

再对人BFS一次，求出走到边界的最少时间。

#include <iostream>

#include <queue>

#include <cstring>

using namespace std;

int n,m,bz[1005][1005],qihuo[1005][1005];

char map[1005][1005];

int h[4][2]={-1,0,1,0,0,-1,0,1};

struct point

{

	int x,y,step;

};

queue<point> q;

void bfs_huo()

{

	int i,j,x,y;

	point t,tt;

	while(!q.empty())

	{

		t=q.front(); q.pop();

		for(i=0;i<4;i++)

		{

			x=t.x+h[i][0]; y=t.y+h[i][1];

			if(x>=0&&x<n&&y>=0&&y<m&&(map[x][y]=='.'||map[x][y]=='J')&&!bz[x][y])

			{

				bz[x][y]=1;

				tt.x=x; tt.y=y; tt.step=t.step+1; qihuo[x][y]=tt.step;

				q.push(tt);

			}

		}

	}

}

int bfs_men(point s)

{

	int i,j,x,y;

	point t,tt;

	while(!q.empty()) q.pop();

	q.push(s); bz[s.x][s.y]=1;

	while(!q.empty())

	{

		t=q.front(); q.pop();

		if(t.x<=0||t.x==n-1||t.y<=0||t.y==m-1) return t.step; //??磬?0

		for(i=0;i<4;i++)

		{

			x=t.x+h[i][0]; y=t.y+h[i][1];

			if(x>=0&&x<n&&y>=0&&y<m&&(map[x][y]=='.'||map[x][y]=='J')&&!bz[x][y])

			{

				if(qihuo[x][y]<=t.step+1)  continue;

				bz[x][y]=1;

				tt.x=x; tt.y=y; tt.step=t.step+1;

				q.push(tt);

			}

		}

	}

	return -1;

}

int main(int argc, char *argv[])

{

	int nn,i,j;

	point s,t;

	cin>>nn;

	while(nn--)

	{

		cin>>n>>m;

		while(!q.empty()) q.pop();

		memset(bz,0,sizeof(bz));

		for(i=0;i<n;i++)

		for(j=0;j<m;j++)

		{

			cin>>map[i][j];   qihuo[i][j]=1000000000; //???

			if(map[i][j]=='J') {s.x=i; s.y=j; s.step=0;}

			if(map[i][j]=='F') {t.x=i; t.y=j; t.step=0; q.push(t); bz[i][j]=1; qihuo[i][j]=0;}

		}

		bfs_huo();

		memset(bz,0,sizeof(bz));

		int ans=bfs_men(s);

		if(ans>=0) cout<<ans+1<<endl;

		else cout<<"IMPOSSIBLE"<<endl;

	}

	return 0;

}