hdu 1242 利用优先队列进行广搜

这题题目不难，代码也不长，花了一个小时，主要是之前没用过结构体的构造函数，比较函数与优先队列

/*
 * hdu1242/linux.cpp
 * Created on: 2011-9-4
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;

typedef struct Node {
    int x, y;
    int time;
    Node(int xx, int yy, int tt) {
        x = xx;
        y = yy;
        time = tt;
    }
    friend bool operator<(const Node &n1, const Node &n2) {
        return n1.time > n2.time;
    }
} Node;

const int MAXN = 205;
const int move[4][2] = { { 1, 0 }, { 0, 1 }, { -1, 0 }, { 0, -1 } };
char map[MAXN][MAXN];
bool visited[MAXN][MAXN];
int M, N;

int work() {
    int sx, sy, xx, yy;
    priority_queue<Node> pq;
    memset(map, '#', sizeof(map));
    memset(visited, false, sizeof(visited));
    for (int i = 1; i <= M; i++) {
        getchar();
        for (int j = 1; j <= N; j++) {
            map[i][j] = getchar();
            if (map[i][j] == 'a') {
                sx = i;
                sy = j;
            }
        }
    }
    pq.push(Node(sx, sy, 0));
    visited[sx][sy] = true;
    while (!pq.empty()) {
        Node cur = pq.top();
        pq.pop();
        for (int i = 0; i < 4; i++) {
            xx = cur.x + move[i][0];
            yy = cur.y + move[i][1];
            if (visited[xx][yy]) {
                continue;
            }
            if (map[xx][yy] == '.') {
                pq.push(Node(xx, yy, cur.time + 1));
            } else if (map[xx][yy] == 'x') {
                pq.push(Node(xx, yy, cur.time + 2));
            } else if (map[xx][yy] == 'r') {
                return cur.time + 1;
            }
            visited[xx][yy] = true;
        }
    }
    return -1;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int ans;
    while (scanf("%d%d", &M, &N) == 2) {
        ans = work();
        if (ans == -1) {
            puts("Poor ANGEL has to stay in the prison all his life.");
        } else {
            printf("%d\n", ans);
        }
    }
    return 0;
}