HDU 1026 Ignatius and the Princess I（带路径的BFS）

http://acm.hdu.edu.cn/showproblem.php?pid=1026

题意:给出一个迷宫，求出到终点的最短时间路径。

这道题目在迷宫上有怪物，不同HP的怪物会损耗不同的时间，这个时候可以用优先队列，每次让时间最短的出队列。由于最后还需要输出路径，所以需要设置一个数组来保存路径。

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstring>
 4 using namespace std;
 5 
 6 const int maxn = 100 + 5;
 7 
 8 struct node
 9 {
10     int x, y;
11     int time;
12     friend bool operator < ( node a, node b)   //重载<号
13     {
14         return b.time<a.time;
15     }
16 };
17 
18 char map[maxn][maxn];
19 int visited[maxn][maxn];
20 int path[maxn][maxn];
21 int  n, m;
22 int d[][2] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } };
23 
24 int bfs()
25 {    
26     node q,now;
27     priority_queue<node> p;
28     q.x = 0;
29     q.y = 0;
30     q.time = 0;
31     p.push(q);
32     while (!p.empty())
33     {
34         q = p.top();
35         p.pop();
36         if (q.x == n - 1 && q.y == m - 1)  return q.time;
37         for (int i = 0; i < 4; i++)
38         {
39             int xx = q.x + d[i][0];
40             int yy = q.y + d[i][1];
41             now.x = xx;
42             now.y = yy;
43             now.time = q.time;
44             if (xx >= 0 && xx < n && yy >= 0 && yy < m && map[xx][yy]!='X' && !visited[xx][yy])
45             {
46                 if (map[xx][yy] == '.')   now.time++;
47                 else now.time =now.time+ (map[xx][yy] - '0'+1);
48                 visited[xx][yy] = 1;
49                 path[xx][yy] = i + 1;     //记录路径
50                 p.push(now);
51             }
52         }
53     }
54     return -1;
55 }
56 
57 int temp;
58 
59 void print(int x, int y)
60 {
61     int xx, yy;
62     if (path[x][y] == 0)   return;
63     xx = x - d[path[x][y] - 1][0];   //寻找第一个路径点
64     yy = y - d[path[x][y] - 1][1];
65     print(xx, yy);
66     printf("%ds:(%d,%d)->(%d,%d)\n", temp++, xx, yy, x, y);
67     if (map[x][y] <= '9' && map[x][y] >= '0')
68     {
69         int m = map[x][y] - '0';
70         while (m--)    printf("%ds:FIGHT AT (%d,%d)\n", temp++, x, y);
71     }
72 }
73 
74 int main()
75 {
76     while (cin >> n >> m && (n||m))
77     {
78         memset(visited, 0, sizeof(visited));
79         memset(path, 0, sizeof(path));
80         for (int i = 0; i < n;i++)
81         for (int j = 0; j < m; j++)
82             cin >> map[i][j];
83         visited[0][0] = 1;
84         int ans=bfs();
85         if (ans == -1)  cout << "God please help our poor hero." << endl;
86         else
87         {
88             cout << "It takes " << ans << " seconds to reach the target position, let me show you the way." << endl;
89             temp = 1;
90             print(n - 1,m - 1);
91         }
92         cout << "FINISH" << endl;
93     }
94     return 0;
95 }

 

posted @ 2016-12-31 13:34  Kayden_Cheung  阅读(168)  评论(0)  编辑  收藏  举报
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