题意：给一个地图，'x'走一步代价为2，'.'走一步代价为1，求从s到t的最小代价。裸优先队列。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 17 using namespace std; 18 19 #define mem0(a) memset(a, 0, sizeof(a)) 20 #define lson l, m, rt << 1 21 #define rson m + 1, r, rt << 1 | 1 22 #define define_m int m = (l + r) >> 1 23 #define Rep(a, b) for(int a = 0; a < b; a++) 24 #define lowbit(x) ((x) & (-(x))) 25 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {} 26 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 27 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 28 29 typedef double db; 30 typedef long long LL; 31 typedef pair<int, int> pii; 32 typedef multiset<int> msi; 33 typedef multiset<int>::iterator msii; 34 typedef set<int> si; 35 typedef set<int>::iterator sii; 36 typedef vector<int> vi; 37 38 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1}; 39 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1}; 40 const int maxn = 1e5 + 7; 41 const int maxm = 1e5 + 7; 42 const int maxv = 1e7 + 7; 43 const int MD = 1e9 +7; 44 const int INF = 1e9 + 7; 45 const double PI = acos(-1.0); 46 const double eps = 1e-10; 47 48 struct Node { 49 int x, y, cost; 50 bool operator < (const Node &a) const { 51 return cost > a.cost; 52 } 53 constructInt3(Node, x, y, cost); 54 }; 55 56 priority_queue<Node> Q; 57 58 int n, m; 59 char s[220][220]; 60 61 void BFS(int x1, int y1, int x2, int y2) { 62 while (!Q.empty()) Q.pop(); 63 Q.push(Node(x1, y1, 0)); 64 s[x1][y1] = '*'; 65 while (!Q.empty()) { 66 Node top = Q.top(); Q.pop(); 67 if (top.x == x2 && top.y == y2) { 68 cout << top.cost << endl; 69 return ; 70 } 71 for (int i = 0; i < 4; i++) { 72 int x = top.x + dx[i], y = top.y + dy[i]; 73 if (x >= 0 && x < n && y >= 0 && y < m && (s[x][y] == '.' || s[x][y] == 'x')) { 74 Q.push(Node(x, y, top.cost + (s[x][y] == '.'? 1 : 2))); 75 s[x][y] = '*'; 76 } 77 } 78 } 79 puts("Poor ANGEL has to stay in the prison all his life."); 80 } 81 82 int main() { 83 //freopen("in.txt", "r", stdin); 84 while (cin >> n >> m) { 85 int x1, x2, y1, y2; 86 for (int i = 0; i < n; i++) { 87 scanf("%s", s + i); 88 for (int j = 0; j < m; j++) { 89 if (s[i][j] == 'r') { 90 x1 = i; 91 y1 = j; 92 s[i][j] = '.'; 93 } 94 if (s[i][j] == 'a') { 95 x2 = i; 96 y2 = j; 97 s[i][j] = '.'; 98 } 99 } 100 } 101 BFS(x1, y1, x2, y2); 102 } 103 return 0; 104 }