uvaoj 11624 - Fire!

题意：一个人从迷宫里往外逃，迷宫里面有火，每秒往周围延伸一格。障碍物人和火都不会到达。问人逃出去最短时间是多少

思路：BFS

代码如下：

#include
#include
#include
#include
#include
using namespace std;

struct pos
{
    int x,y;
    pos(int x=0,int y=0):x(x),y(y){}
};
int r,c;

bool legel(pos t)
{
    if(t.x<0)return 0;
    if(t.y<0)return 0;
    if(t.x>=r)return 0;
    if(t.y>=c)return 0;
    return 1;
}

int dir[4][2]={-1,0,0,-1,1,0,0,1};
vector sav;
char maze[1005][1005];
int ftime[1005][1005];//着火时间
int rtime[1005][1005];//人到达时间

void bfs(int x,int y)//计算着火时间
{
    queue s;
    s.push(pos(x,y));
    while(!s.empty())
    {
        pos now=s.front();
        s.pop();
        for(int d=0;d<4;++d)
        {
            pos tmp=pos(now.x+dir[d][0],now.y+dir[d][1]);
            if(!legel(tmp))continue;
            if(maze[tmp.x][tmp.y]=='#')continue;
            if(ftime[tmp.x][tmp.y]>(ftime[now.x][now.y]+1)||ftime[tmp.x][tmp.y]==-1)
            {
                ftime[tmp.x][tmp.y]=ftime[now.x][now.y]+1;
                s.push(tmp);
            }
        }
    }
}

void solve(int sx,int sy)
{
    queue q;
    q.push(pos(sx,sy));
    rtime[sx][sy]=0;
    while(!q.empty())
    {
        pos now=q.front();
        q.pop();
        for(int d=0;d<4;++d)
        {
            pos tmp=pos(now.x+dir[d][0],now.y+dir[d][1]);
            if(!legel(tmp))continue;
            if(maze[tmp.x][tmp.y]=='#')continue;
            int tt=rtime[now.x][now.y]+1;
            if(ftime[tmp.x][tmp.y]<=tt&&ftime[tmp.x][tmp.y]!=-1)continue;
            if(rtime[tmp.x][tmp.y]>tt||rtime[tmp.x][tmp.y]==-1)
            {
                rtime[tmp.x][tmp.y]=tt;
                q.push(tmp);
            }
        }
    }
    int ans=1000000;
    for(int i=0;i