The 2019 Asia Nanchang First Round Online Programming Contest(B,E)

B. Fire-Fighting Hero

Meaning of the questions: Maximum shortest of firefighters and a team of more competition, comparing the maximum shortest of all places, the final value of the firefighters to take the 1 / C, seek the victory of the party is. I have not read the correct meaning of the title (guilt).

Thinking: FIG topic - single-source shortest path: adding a vertex, the connecting points of the respective fire fighting team is located, the path lengths are set to 0, provided the source vertex, i.e., into a single-source shortest path problem. Dijkstra's algorithm can be determined using twice the maximum two shortest paths. When comparing the fire fighting team multiplied by T score compares to avoid the operation. (answer)

AC Code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 2e6+5;
 5 const ll INF = 1e18 + 10;
 6 typedef pair<ll,int> P;
 7 int n , m;
 8 struct edge{
 9     int to;
10     ll cost;
11 }es[maxn];
12 vector <edge> G[maxn];
13 ll d[maxn];
14 ll dijkstra(int s)
15 {
16     priority_queue<P,vector<P>,greater<P> > que;
17     fill(d,d+n+1,INF);
18     d[s] = 0;
19     que.push(P(0,s));
20     while(!que.empty())
21     {
22         P p= que.top();que.pop();
23         int v = p.second;
24         if(d[v] < p.first) continue;
25         for(int i = 0;i < G[v].size();i++)
26         {
27             edge e = G[v][i];;
28             if(d[e.to] > d[v] + e.cost)
29             {
30 
31                 d[e.to]= d[v] + e.cost;
32                 que.push(P(d[e.to],e.to));
33             }
34         }
35     }
36     ll a = -INF;
37     for(int i = 1;i <= n;i++){
38         a = max(a, d[i]);
39     }
40     return a;
41 }
42 ll a[maxn];
43 int  cnt;
44 void add(int a, int b, ll c)
45 {
46     es[cnt].to = b;
47     es[cnt].cost = c;
48     G[a].push_back(es[cnt++]);
49     es[cnt].to = a;
50     es[cnt].cost = c;
51     G[b].push_back(es[cnt++]);
52 }
53 int main()
54 {
55     std::ios::sync_with_stdio(false);
56     int t, S, k;
57     ll C;
58     cin >> t;
59     while(t--)
60     {
61         cin >> n >> m >> S >> k >> C;
62         int u,v,w;
63         cnt = 0;
64         for(int i = 0;i <= n;i++)G[i].clear();
65         for(int i = 1;i <= k;i++) cin >> a[i];
66         for(int i = 0;i < m;i++)
67         {
68             cin >> u >> v >> w;
69             add(u, v, w);
70         }
71         ll z = dijkstra(S);
72        // cout << z << " = z" << endl;
73         for(int i = 1;i <= k;i++)
74             add(0,a[i],0);
75         ll x = dijkstra(0);
76         //cout << x << " " << endl;
77         cout<<((z <= x * C) ? z : x) <<endl;
78     }
79     return 0;
80 }

View Code

 

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E. Magic Master

Thinking: I never thought that simulated violence on the line, like a long time, and a look ignorant force.

AC Code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int t;
 6     scanf("%d",&t);
 7     while(t--){
 8         int n, m, q;
 9         scanf("%d%d%d",&n, &m, &q);
10         deque<int>  dq;
11         for(int i = n;i > 0;i--)
12         {
13             dq.push_front(i);
14             if(i == 1)break;
15             for(int j = 1;j <= m;j++)
16             {
17                 int x = dq.back();
18                 dq.pop_back();
19                 dq.push_front(x);
20             }
21         }
22         while(q--)
23         {
24             int a;
25             scanf("%d",&a);
26             printf("%d\n",dq[a - 1]);
27         }
28     }
29     return 0;
30 }

View Code

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 H. The Nth Item

To be completed