UVA

题目大意：有n个城市m条航线。给出每条航线的出发地，目的地，座位数，起飞时间和到达时间(所给形式为HHMM。记得转化)，再给出城市A和B。和到达城市B的最晚时间。如今问一天内最多有多少人能从A飞到B，能够在其它城市中转

解题思路：将飞机票拆点，拆成i–>i ＋ m,容量为座位数。
接着推断一下。航线之间的连线
假设航线的起点是A的话，那么就和超级源点相连，容量为INF
假设航线的终点是B且到达时间小于等于最晚时间。那么连线，容量为INF
假设航线i的终点和航线j的起点同样。且航线i的到达时间+30<=航线j的起始时间，那么连线。容量为INF

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <iostream>
using namespace std;
#define N 10010
#define INF 0x3f3f3f3f

struct 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<Edge> edges;
    vector<int> 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<int> 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 160
int n, m, source, sink, Time;
int num[S];
map<string, int> 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;
}