ZOJ-1649 Rescue---BFS+优先队列

题目链接：

https://vjudge.net/problem/ZOJ-1649

题目大意：

天使的朋友要去救天使，a是天使，r 是朋友，x是卫兵。每走一步需要时间1，打倒卫兵需要另外的时间1，问救到天使所用的最少时间。注意存在救不到的情况。

思路：

BFS搜索，由于打倒卫兵时间为2，所以用BFS+优先队列做，每次出队时间最少的拓展，每个格子只走一次才是最优解

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<sstream>
#include<functional>
using namespace std;
typedef long long ll;
const int maxn = 2e2 + 10;
const int INF = 1e9 + 7;
int T, n, m, cases;
int dir[][2] = {1,0,0,1,-1,0,0,-1};
struct node
{
    int x, y, time;
    bool operator <(const node& a)const
    {
        return time > a.time;
    }
    node(){}
    node(int x, int y, int time):x(x), y(y), time(time){}
};
char Map[maxn][maxn];
bool vis[maxn][maxn];
bool judge(int x, int y)
{
    return (x >= 0 && x < n && y >= 0 && y < m && !vis[x][y] && Map[x][y] != '#');
}
void bfs(int x, int y)
{
    memset(vis, 0, sizeof(vis));
    priority_queue<node>q;
    q.push(node(x, y, 0));
    vis[x][y] = 1;
    while(!q.empty())
    {
        node now = q.top();
        q.pop();
        if(Map[now.x][now.y] == 'r')
        {
            cout<<now.time<<endl;
            return;
        }
        for(int i = 0; i < 4; i++)
        {
            node next = now;
            next.x += dir[i][0];
            next.y += dir[i][1];
            if(judge(next.x, next.y))
            {
                vis[next.x][next.y] = 1;
                next.time++;
                if(Map[next.x][next.y] == 'x')next.time++;
                q.push(next);
            }
        }
    }
    printf("Poor ANGEL has to stay in the prison all his life.
");
    return;
}
int main()
{
    while(cin >> n >> m)
    {
        int sx, sy;
        for(int i = 0; i < n; i++)
        {
            cin >> Map[i];
            for(int j = 0; j < m; j++)if(Map[i][j] == 'a')sx = i, sy = j;
        }
        bfs(sx, sy);
    }
    return 0;
}