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。

 

 1 #include <iostream>
 2 #include <cstring>
 3 #include <queue>
 4 
 5 using namespace std;
 6 const int M = 10;
 7 int map[M][M];
 8 int graph[M][M];
 9 struct node
10 {
11     int x,y;
12     int step;
13     int time;
14 };
15 
16 node now,nn,next1;
17 queue <struct node> q;
18 int n,m;
19 int dire[4][2] = {{-1,0},{0,-1},{1,0},{0,1}};
20 
21 int bfs()
22 {
23     while (!q.empty())
24     {
25         now = q.front();
26         q.pop();
27         for (int i = 0;i < 4;i++)
28         {
29             int x = now.x + dire[i][0];
30             int y = now.y + dire[i][1];
31             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
32             {                
33                 next1.x = x;
34                 next1.y = y;
35                 if (map[x][y] == 3)
36                 {
37                     return now.step+1;
38                 }
39                 if (map[x][y] == 4)
40                 {
41                     next1.time = 6;
42                 }
43                 else next1.time = now.time - 1;
44                 next1.step = now.step+1;
45                 if (next1.time > graph[x][y])    //加入队列条件
46                 {
47                     graph[x][y] = next1.time;
48                     q.push(next1);
49                 }
50             }
51         }
52     }
53     return -1;
54 }
55 int main()
56 {
57     int t;
58     cin >> t;
59     while (t--)
60     {
61         cin >> n >> m;
62         while (!q.empty())
63             q.pop();
64         for (int i = 0;i < n;i++)
65         {
66             for (int j = 0;j < m;j++)
67             {
68                 cin >> map[i][j];
69                 graph[i][j] = 0;
70                 if (map[i][j] == 2)
71                 {
72                     nn.x = i;
73                     nn.y = j;
74                     nn.step = 0;
75                     nn.time = 6;
76                     q.push(nn);
77                 }
78             }
79         }
80         cout << bfs() << endl;
81     }
82     return 0;
83 }

 

posted @ 2015-08-19 19:58  ediszhao  阅读(175)  评论(0)  编辑  收藏  举报