[hdu1242]优先队列

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。

 #pragma comment(linker, "/STACK:10240000,10240000")
 
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 
 using namespace std;
 
 #define mem0(a) memset(a, 0, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define define_m int m = (l + r) >> 1
 #define Rep(a, b) for(int a = 0; a < b; a++)
 #define lowbit(x) ((x) & (-(x)))
 #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) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 
 typedef double db;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef multiset<int> msi;
 typedef multiset<int>::iterator msii;
 typedef set<int> si;
 typedef set<int>::iterator sii;
 typedef vector<int> vi;
 
 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {, -, , , -, , , -};
 const int maxn = 1e5 + ;
 const int maxm = 1e5 + ;
 const int maxv = 1e7 + ;
 const int MD = 1e9 +;
 const int INF = 1e9 + ;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 
 struct Node {
     int x, y, cost;
     bool operator < (const Node &a) const {
         return cost > a.cost;
     }
     constructInt3(Node, x, y, cost);
 };
 
 priority_queue<Node> Q;
 
 int n, m;
 char s[][];
 
 void BFS(int x1, int y1, int x2, int y2) {
     while (!Q.empty()) Q.pop();
     Q.push(Node(x1, y1, ));
     s[x1][y1] = '*';
     while (!Q.empty()) {
         Node top = Q.top(); Q.pop();
         if (top.x == x2 && top.y == y2) {
             cout << top.cost << endl;
             return ;
         }
         for (int i = ; i < ; i++) {
             int x = top.x + dx[i], y = top.y + dy[i];
             if (x >=  && x < n && y >=  && y < m && (s[x][y] == '.' || s[x][y] == 'x')) {
                 Q.push(Node(x, y, top.cost + (s[x][y] == '.'?  : )));
                 s[x][y] = '*';
             }
         }
     }
     puts("Poor ANGEL has to stay in the prison all his life.");
 }
 
 int main() {
     //freopen("in.txt", "r", stdin);
     while (cin >> n >> m) {
         int x1, x2, y1, y2;
         for (int i = ; i < n; i++) {
             scanf("%s", s + i);
             for (int j = ; j < m; j++) {
                 if (s[i][j] == 'r') {
                     x1 = i;
                     y1 = j;
                     s[i][j] = '.';
                 }
                 if (s[i][j] == 'a') {
                     x2 = i;
                     y2 = j;
                     s[i][j] = '.';
                 }
             }
         }
         BFS(x1, y1, x2, y2);
     }
     return ;
 }