UVA 11624 Fire!(二次BFS)

先對火BFS一次，求出每個點的最小著火時間。

再對人BFS一次，求出走到邊界的最少時間。

#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;
}