hdoj1072 Nightmare bfs

题意：在一个地图里逃亡，2是起点，3是终点，1是路，0是墙，逃亡者携带一个炸弹，6分钟就会炸，要在6分钟前到达4可以重制时间，问是否能逃亡，若能则输出最小值

我的思路：bfs在5步内是否存在3，存在则输出退出。记录到达每一点剩余时间，如果再次到达某点的剩余时间大于原剩余时间则更新，加入队列。

 #include <iostream>
 #include <cstring>
 #include <queue>

 using namespace std;
 const int M = ;
 int map[M][M];
 int graph[M][M];
 struct node
 {
     int x,y;
     int step;
     int time;
 };

 node now,nn,next1;
 queue <struct node> q;
 int n,m;
 int dire[][] = {{-,},{,-},{,},{,}};

 int bfs()
 {
     while (!q.empty())
     {
         now = q.front();
         q.pop();
         for (int i = ;i < ;i++)
         {
             int x = now.x + dire[i][];
             int y = now.y + dire[i][];
             if (x>= && x<n && y>= && y<m && map[x][y] !=  && map[x][y]!= && now.time->)  //踩点到到达4也会爆炸
             {
                 next1.x = x;
                 next1.y = y;
                 if (map[x][y] == )
                 {
                     return now.step+;
                 }
                 if (map[x][y] == )
                 {
                     next1.time = ;
                 }
                 else next1.time = now.time - ;
                 next1.step = now.step+;
                 if (next1.time > graph[x][y])    //加入队列条件
                 {
                     graph[x][y] = next1.time;
                     q.push(next1);
                 }
             }
         }
     }
     return -;
 }
 int main()
 {
     int t;
     cin >> t;
     while (t--)
     {
         cin >> n >> m;
         while (!q.empty())
             q.pop();
         for (int i = ;i < n;i++)
         {
             for (int j = ;j < m;j++)
             {
                 cin >> map[i][j];
                 graph[i][j] = ;
                 if (map[i][j] == )
                 {
                     nn.x = i;
                     nn.y = j;
                     nn.step = ;
                     nn.time = ;
                     q.push(nn);
                 }
             }
         }
         cout << bfs() << endl;
     }
     return ;
 }