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 = 10;
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[4][2] = {{-1,0},{0,-1},{1,0},{0,1}};

int bfs()
{
    while (!q.empty())
    {
        now = q.front();
        q.pop();
        for (int i = 0;i < 4;i++)
        {
            int x = now.x + dire[i][0];
            int y = now.y + dire[i][1];
            if (x>=0 && x<n && y>=0 && y<m && map[x][y] != 0 && map[x][y]!=2 && now.time-1>0)  //踩点到到达4也会爆炸
            {                
                next1.x = x;
                next1.y = y;
                if (map[x][y] == 3)
                {
                    return now.step+1;
                }
                if (map[x][y] == 4)
                {
                    next1.time = 6;
                }
                else next1.time = now.time - 1;
                next1.step = now.step+1;
                if (next1.time > graph[x][y])    //加入队列条件
                {
                    graph[x][y] = next1.time;
                    q.push(next1);
                }
            }
        }
    }
    return -1;
}
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        cin >> n >> m;
        while (!q.empty())
            q.pop();
        for (int i = 0;i < n;i++)
        {
            for (int j = 0;j < m;j++)
            {
                cin >> map[i][j];
                graph[i][j] = 0;
                if (map[i][j] == 2)
                {
                    nn.x = i;
                    nn.y = j;
                    nn.step = 0;
                    nn.time = 6;
                    q.push(nn);
                }
            }
        }
        cout << bfs() << endl;
    }
    return 0;
}