[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[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
const int maxn = 1e5 + 7;
const int maxm = 1e5 + 7;
const int maxv = 1e7 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-10;

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[220][220];

void BFS(int x1, int y1, int x2, int y2) {
    while (!Q.empty()) Q.pop();
    Q.push(Node(x1, y1, 0));
    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 = 0; i < 4; i++) {
            int x = top.x + dx[i], y = top.y + dy[i];
            if (x >= 0 && x < n && y >= 0 && y < m && (s[x][y] == '.' || s[x][y] == 'x')) {
                Q.push(Node(x, y, top.cost + (s[x][y] == '.'? 1 : 2)));
                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 = 0; i < n; i++) {
            scanf("%s", s + i);
            for (int j = 0; 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 0;
}