題意：

‘S‘ : 起點

‘T‘ : 終點

‘#‘ : 毒氣室

‘B‘ :氧氣

‘P‘:不消耗步數

每次經過毒氣室需要一瓶氧氣，氧氣可以重復獲得，但只能帶五瓶氧氣，問最少步數

solution：

HINT:多維狀態判重，多一維攜帶氧氣瓶數量

沒帶氧氣瓶的時候不能走毒氣室#

攜帶超過5個跳過氧氣B

相似題目：UVA816 Abbott‘s Revenge這題多一維方向

  1 #include<bits/stdc++.h>
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<algorithm>
  5
 
 #include<vector>
  6 #include<queue>
  7 #include<cstring>
  8 #define fi first
  9 #define se second
 10 #define mp make_pair
 11 #define pb push_back
 12 #define pw(x) (1ll << (x))
 13 #define sz(x) ((int)(x).size())
 14 #define all(x) (x).begin(),(x).end()
 15 #define rep(i,l,r) for(int i=(l);i<(r);++i)
 16
 
 #define per(i,l,r) for(int i=(r)-1;i>=(l);--i)
 17 #define maxn 500005
 18 #define eps 1e-9
 19 #define PIE acos(-1)
 20 #define dd(x) cout << #x << " = " << (x) << ", "
 21 #define de(x) cout << #x << " = " << (x) << "\n"
 22 #define endl "\n"
 23 #define
 
 INF 0x3f3f3f3f
 24 using namespace std;
 25 typedef double db;
 26 typedef long long LL;
 27 typedef vector<int> vi;
 28 typedef pair<int, int> pii;
 29 //----------------------
 30 int n,m;
 31 char pic[105][105];
 32 bool vis[105][105][6];
 33 const int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
 34 struct Node{
 35     int x;
 36     int y;
 37     int d=0;
 38     int cnt=0; 
 39     bool operator<(const Node& a)const{
 40         return d>a.d;
 41     }
 42 }start,u,v;
 43 int cnt;
 44 bool check(Node v)
 45 {
 46     return v.x>=0&&v.x<n&&v.y>=0&&v.y<m;
 47 }
 48 int bfs(int x,int y)//把#當路障,找氧氣瓶 
 49 {
 50     memset(vis,0,sizeof(vis));
 51     priority_queue<Node>q;
 52     vis[x][y][0]=1;
 53     start.x=x;start.y=y;start.d=0;start.cnt=0;
 54     q.push(start);
 55     while(!q.empty())
 56     {
 57         u=q.top();q.pop();
 58         if(pic[u.x][u.y]==‘T‘){return u.d;}
 59         rep(i,0,4){
 60             v=u;
 61             v.x+=dir[i][0];
 62             v.y+=dir[i][1];
 63             if(!check(v))continue;
 64             if(pic[v.x][v.y]==‘#‘){
 65                 if(v.cnt>=1)v.cnt--,v.d++;
 66                 else continue;
 67             }
 68             else if(pic[v.x][v.y]==‘B‘){
 69                 if(v.cnt>=5)continue;
 70                 else v.cnt++;
 71             }
 72             else if(pic[v.x][v.y]==‘P‘)v.d--;
 73             v.d++;
 74             if(vis[v.x][v.y][v.cnt])continue;
 75             vis[v.x][v.y][v.cnt]=1;
 76 //             dd(v.x+1);dd(v.y+1);dd(pic[v.x][v.y]);dd(v.cnt);de(v.d);
 77             q.push(v);
 78         }
 79     }
 80     return INF;
 81 }
 82 int main()
 83 {
 84 //    ifstream cin("in.txt");
 85     while(cin>>n>>m,n+m){
 86         int ans=INF;
 87         int ok=0;
 88         rep(i,0,n)rep(j,0,m){
 89             cin>>pic[i][j];
 90         }
 91         rep(i,0,n)rep(j,0,m)if(pic[i][j]==‘S‘)
 92         {
 93             ans=bfs(i,j);
 94             goto here;
 95         }
 96         here:;
 97         if(ans!=INF)printf("%d\n",ans);
 98         else puts("-1");
 99     }
100     return 0;
101 }

BFS，優先隊列優化