UVA 11624

题目链接：https://cn.vjudge.net/problem/UVA-11624

题意：

给出一个 n×m 的矩阵，上面有的格子能走，有的格子是墙不能走。

有若干个点是火源，火每分钟都往上下左右蔓延一格（不能越墙）。又给出一个点是Joe的出发点，他只要能在不碰到火的前提下走出该矩阵，就算逃生成功，要求最小的逃生时间。

题解：

显然可以把火和人一起扔进队列里BFS，只要初始状态的时候火比人先进队，那么往后在BFS的每一层内永远都是火比人先进队。

然后，book数组正好也可以用来被火给标记掉，可以使得当前时间下人不能走火走过格子。

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1003;
const int INF = 0x3f3f3f3f;
const int dx[4] = { 1,0,-1,0 };
const int dy[4] = { 0,1,0,-1 };

int n, m;
char tmp[maxn];
int mp[maxn][maxn];

struct Node {
   bool fire;
   int x, y;
   int step;
   Node() {}
   Node(bool _fire, int _x, int _y, int _step) {
       fire = _fire;
       x = _x, y = _y;
       step = _step;
   }
}joe, fire[maxn*maxn];
int firecnt;

int vis[maxn][maxn];
queue<Node> Q;
inline bool border(Node o) {
   return o.x == 1 || o.x == n || o.y == 1 || o.y == m;
}
int bfs()
{
   memset(vis, 0, sizeof(vis));
   while (!Q.empty()) Q.pop();

   for (int i = 1; i <= firecnt; i++)
   {
       fire[i].fire = 1;
       fire[i].step = 0;
       Q.push(fire[i]);
       vis[fire[i].x][fire[i].y] = 1;
   }

   joe.step = joe.fire = 0;
   Q.push(joe);
   vis[joe.x][joe.y] = 1;

   while (!Q.empty())
   {
       Node now = Q.front(); Q.pop();
       if (!now.fire && border(now)) return now.step;
       for (int k = 0; k < 4; k++)
       {
           Node nxt = Node(now.fire, now.x + dx[k], now.y + dy[k], now.step + 1);
           if (mp[nxt.x][nxt.y]) continue;
           if (vis[nxt.x][nxt.y]) continue;
           vis[nxt.x][nxt.y] = 1;
           Q.push(nxt);
       }
   }
   return -1;
}

int main()
{
   int T;
   cin >> T;
   while (T--)
   {
       scanf("%d%d", &n, &m);
       memset(mp, 1, sizeof(mp));
       firecnt = 0;
       for (int i = 1; i <= n; i++)
       {
           scanf("%s", tmp + 1);
           for (int j = 1; j <= m; j++)
           {
               if (tmp[j] == '#') mp[i][j] = 1;
               if (tmp[j] == '.') mp[i][j] = 0;
               if (tmp[j] == 'J') joe.x = i, joe.y = j, mp[i][j] = 0;
               if (tmp[j] == 'F') firecnt++, fire[firecnt].x = i, fire[firecnt].y = j, mp[i][j] = 0;
           }
       }
       int ans = bfs();
       if (ans == -1) printf("IMPOSSIBLE
");
       else printf("%d
", ans + 1);
   }
}