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) ", 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) ", 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 }