BFS，优先队列优化

题意：

'S' : 起点

'T' : 终点

'#' : 毒气室

'B' :氧气

'P':不消耗步数

每次经过毒气室需要一瓶氧气，氧气可以重复获得，但只能带五瓶氧气，问最少步数

solution：

HINT:多维状态判重，多一维携带氧气瓶数量

没带氧气瓶的时候不能走毒气室#

携带超过5个跳过氧气B

相似题目：UVA816 Abbott's Revenge这题多一维方向

 #include<bits/stdc++.h>
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #include<cstring>
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 #define pw(x) (1ll << (x))
 #define sz(x) ((int)(x).size())
 #define all(x) (x).begin(),(x).end()
 #define rep(i,l,r) for(int i=(l);i<(r);++i)
 #define per(i,l,r) for(int i=(r)-1;i>=(l);--i)
 #define maxn 500005
 #define eps 1e-9
 #define PIE acos(-1)
 #define dd(x) cout << #x << " = " << (x) << ", "
 #define de(x) cout << #x << " = " << (x) << "\n"
 #define endl "\n"
 #define INF 0x3f3f3f3f
 using namespace std;
 typedef double db;
 typedef long long LL;
 typedef vector<int> vi;
 typedef pair<int, int> pii;
 //----------------------
 int n,m;
 char pic[][];
 bool vis[][][];
 const int dir[][]={{-,},{,},{,},{,-}};
 struct Node{
     int x;
     int y;
     int d=;
     int cnt=;
     bool operator<(const Node& a)const{
         return d>a.d;
     }
 }start,u,v;
 int cnt;
 bool check(Node v)
 {
     return v.x>=&&v.x<n&&v.y>=&&v.y<m;
 }
 int bfs(int x,int y)//把#当路障,找氧气瓶
 {
     memset(vis,,sizeof(vis));
     priority_queue<Node>q;
     vis[x][y][]=;
     start.x=x;start.y=y;start.d=;start.cnt=;
     q.push(start);
     while(!q.empty())
     {
         u=q.top();q.pop();
         if(pic[u.x][u.y]=='T'){return u.d;}
         rep(i,,){
             v=u;
             v.x+=dir[i][];
             v.y+=dir[i][];
             if(!check(v))continue;
             if(pic[v.x][v.y]=='#'){
                 if(v.cnt>=)v.cnt--,v.d++;
                 else continue;
             }
             else if(pic[v.x][v.y]=='B'){
                 if(v.cnt>=)continue;
                 else v.cnt++;
             }
             else if(pic[v.x][v.y]=='P')v.d--;
             v.d++;
             if(vis[v.x][v.y][v.cnt])continue;
             vis[v.x][v.y][v.cnt]=;
 //             dd(v.x+1);dd(v.y+1);dd(pic[v.x][v.y]);dd(v.cnt);de(v.d);
             q.push(v);
         }
     }
     return INF;
 }
 int main()
 {
 //    ifstream cin("in.txt");
     while(cin>>n>>m,n+m){
         int ans=INF;
         int ok=;
         rep(i,,n)rep(j,,m){
             cin>>pic[i][j];
         }
         rep(i,,n)rep(j,,m)if(pic[i][j]=='S')
         {
             ans=bfs(i,j);
             goto here;
         }
         here:;
         if(ans!=INF)printf("%d\n",ans);
         else puts("-1");
     }
     return ;
 }