Sha-1161 Objective: Berlin (maximum stream + Time Series Model)

Sha-1161 Objective: Berlin (maximum stream + Time Series Model)

There are m routes in n cities, which provide the departure location, destination, number of seats, departure time and arrival time of each route (in the form of HHMM, remember to convert ), next, let's talk about City A and B, and the latest time to arrive at City B. Now, we can ask how many people can fly from A to B in A day, and transfer to another city.

Solution: Split the plane ticket into I-> I + m, and set the capacity to the number of seats.
Then judge the connection between the routes.
If the starting point of the route is A, it is connected to the super source point with the capacity of INF.
If the end of the route is B and the arrival time is less than or equal to the latest time, the connection and capacity are INF.
If the endpoint of route I is the same as the start point of Route j, and the arrival time of route I is + 30 <= start time of Route j, the connection is established and the capacity is INF.

#include 
  
   #include 
   
    #include #include 
    
     #include 
     
      #include
      #include 
       
        using namespace std;#define N 10010#define INF 0x3f3f3f3fstruct Edge{ int from, to, cap, flow; Edge() {} Edge(int from, int to, int cap, int flow) : from(from), to(to), cap(cap), flow(flow) {}};struct Dinic{ int n, m, s, t; vector
        
          edges; vector
         
           G[N]; bool vis[N]; int d[N], cur[N]; void init(int n) { this->n = n; for (int i = 0; i <= n; i++) { G[i].clear(); } edges.clear(); } void AddEdge(int from, int to, int cap) { edges.push_back(Edge(from, to, cap, 0)); edges.push_back(Edge(to, from, 0, 0)); int m = edges.size(); G[from].push_back(m - 2); G[to].push_back(m - 1); } bool BFS() { memset(vis, 0, sizeof(vis)); queue
          
            Q; Q.push(s); vis[s] = 1; d[s] = 0; while (!Q.empty()) { int u = Q.front(); Q.pop(); for (int i = 0; i < G[u].size(); i++) { Edge &e = edges[G[u][i]]; if (!vis[e.to] && e.cap > e.flow) { vis[e.to] = true; d[e.to] = d[u] + 1; Q.push(e.to); } } } return vis[t]; } int DFS(int x, int a) { if (x == t || a == 0) return a; int flow = 0, f; for (int i = cur[x]; i < G[x].size(); i++) { Edge &e = edges[G[x][i]]; if (d[x] + 1 == d[e.to] && (f = DFS(e.to, min(a, e.cap - e.flow))) > 0) { e.flow += f; edges[G[x][i] ^ 1].flow -= f; flow += f; a -= f; if (a == 0) break; } } return flow; } int Maxflow(int s, int t) { this->s = s; this->t = t; int flow = 0; while (BFS()) { memset(cur, 0, sizeof(cur)); flow += DFS(s, INF); } return flow; }};Dinic dinic;#define M 5100#define S 160int n, m, source, sink, Time;int num[S];map
           
             Map;struct Node { int u, v, c, s, t;}node[M];int getTime(string T) { int a = (T[0] - '0') * 10 + (T[1] - '0'); int b = (T[2] - '0') * 10 + (T[3] - '0'); return a * 60 + b;}void solve() { Map.clear(); int cnt = 3; string a, b, s, t; cin >> a >> b >> s >> m; Map[a] = 1; Map[b] = 2; Time = getTime(s); memset(num, 0, sizeof(num)); source = 0; sink = 2 * m + 1; dinic.init(sink); for (int i = 1; i <= m; i++) { cin >> a >> b >> node[i].c >> s >> t; if (!Map[a]) Map[a] = cnt++; if (!Map[b]) Map[b] = cnt++; node[i].u = Map[a]; node[i].v = Map[b]; node[i].s = getTime(s); node[i].t = getTime(t); num[node[i].u]++; num[node[i].v]++; dinic.AddEdge(i, i + m, node[i].c); } if (!num[1] || !num[2]) { printf(0); return ; } for (int i = 1; i <= m; i++) { int u = node[i].u, v = node[i].v; if (u == 1) dinic.AddEdge(source, i, INF); if (v == 2 && node[i].t <= Time) dinic.AddEdge(i + m, sink, INF); for (int j = 1; j <= m; j++) { if (i == j) continue; if (v != node[j].u) continue; if (node[i].t + 30 <= node[j].s) dinic.AddEdge(i + m, j, INF); } } int ans = dinic.Maxflow(source, sink); printf(%d, ans);}int main() { while (scanf(%d, &n) != EOF) solve(); return 0;}