难得的一A
与SPFA判负圈的思想类似. 记录一下入队次数就ok.
我想到的一些优化:
可以找到Q中最大的一个当做入队次数的上界.
因为是多次入队,选择SPFA而非dijkstra (dijkstra会多一个logn)
因为我懒, 以上优化并没有实现,因为在这个数据规模下随便怎么写都行啦
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 #include <map>
8 using namespace std;
9 const int MAXN = 1e2 + 20;
10 const int INF = 0x3f3f3f3f;
11
12 map<string, int> num; int numcnt = 0;
13 inline int getnum(string &s)
14 {
15 if(num.find(s) == num.end())
16 return (num[s] = ++numcnt, numcnt);
17 else return num[s];
18 }
19 struct edge
20 {
21 int to, cost;
22 edge(int v = 0, int c = 0) :
23 to(v), cost(c) {}
24 };vector<edge> g[MAXN];
25 struct state
26 {
27 int pos, k, cost;
28 state(int p = 0, int k = 0, int c = 0) :
29 pos(p), k(k), cost(c) {}
30 bool operator >(const state &lhs) const{
31 return cost > lhs.cost;
32 }
33 };
34 int N, M, Q, S, T;
35
36 int f[MAXN][MAXN];
37 bool vis[MAXN][MAXN];
38
39 inline bool tension(const int &st, int &lg)
40 {
41 return st < lg ? (lg = st, true) : false;
42 }
43
44 priority_queue<state, vector<state>, greater<state> > q;
45 void bfs()
46 {
47 f[S][0] = 0, q.push(state(S, 0, 0));
48 while(!q.empty())
49 {
50 state u = q.top(); q.pop();
51 if(u.k > N || vis[u.pos][u.k]) continue;
52 vis[u.pos][u.k] = true;
53 for(int i = 0; i < (int) g[u.pos].size(); i++)
54 {
55 edge &e = g[u.pos][i];
56 if(tension(f[u.pos][u.k] + e.cost, f[e.to][u.k + 1]))
57 q.push(state(e.to, u.k + 1, f[e.to][u.k + 1]));
58 }
59 }
60 }
61
62 void init()
63 {
64 memset(vis, false, sizeof(vis));
65 memset(f, 0x3f, sizeof(f));
66 num.clear(); numcnt = 0;
67 for(int i = 1; i <= N; i++) g[i].clear();
68 }
69
70 int main()
71 {
72 //freopen("11280.out", "w", stdout);
73 int b, t = 0; cin>>b;
74 while(b--)
75 {
76 if(t) puts("");
77 t++; cin>>N;
78 string s; init();
79 for(int i = 1; i <= N; i++) {cin>>s; getnum(s);}
80 cin>>M;
81 for(int i = 1, u, v, c; i <= M; i++)
82 {
83 cin>>s; u = getnum(s);
84 cin>>s; v = getnum(s);
85 cin>>c; g[u].push_back(edge(v, c));
86 }
87 S = num["Calgary"], T = num["Fredericton"];
88 bfs();
89
90 printf("Scenario #%d
", t);
91 cin>>Q;
92 while(Q--)
93 {
94 int tmp; cin>>tmp;
95 int ans = INF;
96 for(int i = 1; i <= tmp + 1; i++)
97 ans = min(ans, f[T][i]);
98 if(ans == INF) puts("No satisfactory flights");
99 else printf("Total cost of flight(s) is $%d
", ans);
100 }
101
102 }
103 return 0;
104 }