UVA11367 / poj 3653 Full Tank?

容易证明，使用优先队列的BFS中，每个状态(或者说阶段)第一次出队时即为该状态的最小代价。

具体在这道题中，当终点T的某个状态第一次被取出时，即为最优答案/

(我不会说我第一次写没有用记忆化然后TLE了)

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <queue>
 4 #include <vector>
 5 #include <cstring>
 6 #include <iostream>
 7 using namespace std;
 8 const int MAXN = 1000 + 20;
 9 const int INF = 0x3f3f3f3f;
10 
11 int N, M, Q;
12 int C, S, T;
13 int cc[MAXN], d[MAXN][110];
14 
15 struct edge
16 {
17     int to, dis;
18     edge(int v = 0, int c = 0) : to(v), dis(c) {}
19 };
20 
21 vector<edge> g[MAXN];
22 
23 struct P
24 {
25     int city, fuel, cost;
26     bool operator >(const P &x) const{
27         return cost > x.cost;
28     }
29 }ans;
30 
31 inline bool solve()
32 {
33     priority_queue<P, vector<P>, greater<P> > q;
34 
35     d[S][0] = 0;
36     P s;
37     s.city = S, s.fuel = 0, s.cost = 0;
38     q.push(s);
39     while(!q.empty())
40     {
41         P p = q.top(); q.pop();
42         if(p.city == T)
43         {
44             ans = p; return true;
45         }
46         if(p.fuel < C && d[p.city][p.fuel + 1] > d[p.city][p.fuel] + cc[p.city]) 
47         {
48             d[p.city][p.fuel + 1] = d[p.city][p.fuel] + cc[p.city];
49             P nxt = p;
50             nxt.fuel++;
51             nxt.cost += cc[p.city];
52             q.push(nxt);
53         }
54 
55         int u = p.city;
56         for(int i = 0; i < (int) g[u].size(); i++)
57         {
58             const edge &e = g[u][i];
59             if(e.dis <= p.fuel && d[e.to][p.fuel - e.dis] > p.cost)
60             {
61                 d[e.to][p.fuel - e.dis] = p.cost;
62                 P nxt = p;
63                 nxt.fuel -= e.dis;
64                 nxt.city = e.to;
65                 q.push(nxt);
66             }
67         }
68     }
69     return false;
70 }
71 
72 int main()
73 {
74     //freopen("11367.txt", "r", stdin);
75     cin>>N>>M;
76     for(int i = 0; i < N; i++)
77         scanf("%d", &cc[i]);
78 
79     int u, v, c;
80     for(int i = 1; i <= M; i++)
81     {
82         scanf("%d%d%d", &u, &v, &c);
83         g[u].push_back(edge(v, c));
84         g[v].push_back(edge(u, c));
85     }    
86 
87     cin>>Q;
88     while(Q--)
89     {
90         cin>>C>>S>>T;
91         memset(d, 0x3f, sizeof(d));
92         if(solve())    cout<<ans.cost<<endl;
93         else puts("impossible");
94     }
95     return 0;
96 }