hdu1242 又又又是逃离迷宫（bfs模板题）

题目链接：http://icpc.njust.edu.cn/Problem/Hdu/1242/

这次的迷宫是有守卫的，杀死一个守卫需要花费1个单位的时间，所以以走的步数为深度，在每一层进行搜索，由于走一步的花费不一定是1，所以我们需要用优先队列寻找最优值。这个题目真是模板题。

代码如下：

 #include<bits/stdc++.h>
 using namespace std;
 typedef unsigned int ui;
 typedef long long ll;
 typedef unsigned long long ull;
 #define pf printf
 #define mem(a,b) memset(a,b,sizeof(a))
 #define prime1 1e9+7
 #define prime2 1e9+9
 #define pi 3.14159265
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define scand(x) scanf("%llf",&x)
 #define f(i,a,b) for(int i=a;i<=b;i++)
 #define scan(a) scanf("%d",&a)
 #define dbg(args) cout<<#args<<":"<<args<<endl;
 #define inf 0x3f3f3f3f
 #define maxn 205
 int n,m,t;
 int sx,sy,tx,ty;
 bool vis[maxn][maxn];
 char Map[maxn][maxn];
 int dir[][]={,,,-,,,-,};
 struct node{
     int x,y,time;
     node(int x,int y,int s):x(x),y(y),time(s){}
     node(){}
     friend operator< (node a,node b)
     {
         return a.time>b.time;//逆定义运算符
     }
 };
 int ans=;
 node cur,nxt;
 void bfs()
 {
     priority_queue<node>q;
     q.push(node(sx,sy,));
     vis[sx][sy]=;
     while(!q.empty())
     {
         cur=q.top();
         q.pop();
         if(cur.x==tx&&cur.y==ty)
         {
             ans=cur.time;
             return;
         }
         f(i,,)
         {
             nxt=cur;
             nxt.x+=dir[i][];
             nxt.y+=dir[i][];
             nxt.time++;
             if(nxt.x<||nxt.x>n||nxt.y<||nxt.y>m||Map[nxt.x][nxt.y]=='#')continue;
             if(vis[nxt.x][nxt.y])continue;
             vis[nxt.x][nxt.y]=;
             if(Map[nxt.x][nxt.y]=='x')nxt.time++;
             q.push(nxt);
         }
     }
 }
 int main()
 {
     //freopen("input.txt","r",stdin);
     //freopen("output.txt","w",stdout);
     std::ios::sync_with_stdio(false);
     while(scanf("%d%d",&n,&m)!=EOF)
     {
         ans=;
         mem(vis,false);
         f(i,,n)
             f(j,,m)
             {
                 scanf(" %c",&Map[i][j]);
                 if(Map[i][j]=='r')sx=i,sy=j;
                 if(Map[i][j]=='a')tx=i,ty=j;
             }
         bfs();
         if(!ans)pf("Poor ANGEL has to stay in the prison all his life.\n");
         else pf("%d\n",ans);
     }

  }