题意：有n个城市，m条航班。已知每条航班的起点和终点，还有每条航班的载客量、出发时间、到达时间。并且要求在任何一个城市（起点、终点除外）都至少要有30分钟的中转时间，求起点到终点的最大客流量。
析：把每个航线看成一个点，然后拆成两个点，然后如果两个航线能够到达，并且时间不超的话，就连一条边，然后加一个源点和汇点，分别向开始和结束城市进行连线，容量无限大。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-3;
const int maxn = 1e4 + 10;
const int maxm = 1e7 + 10;
const int mod = 1000000007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
} struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
int d[maxn];
bool vis[maxn];
int cur[maxn]; void init(int n){
this-> n = n;
for(int i = 0; i < n; ++i) G[i].cl;
edges.cl;
} void addEdge(int from, int to, int cap){
edges.pb((Edge){from, to, cap, 0});
edges.pb((Edge){to, from, 0, 0});
m = edges.sz;
G[from].pb(m - 2);
G[to].pb(m - 1);
} bool bfs(){
ms(vis, 0); vis[s] = 1; d[s] = 0;
queue<int> q; q.push(s); while(!q.empty()){
int u = q.front(); q.pop();
for(int i = 0; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(!vis[e.to] && e.cap > e.flow){
vis[e.to] = 1;
d[e.to] = d[u] + 1;
q.push(e.to);
}
}
}
return vis[t];
} int dfs(int u, int a){
if(u == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[u]; i < G[u].sz; ++i){
Edge &e = edges[G[u][i]];
if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
e.flow += f;
edges[G[u][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()){ ms(cur, 0); flow += dfs(s, INF); }
return flow;
}
}; Dinic dinic;
map<string, int> mp; int ID(const string &s){
if(mp.count(s)) return mp[s];
return mp[s] = mp.sz;
} struct Fly{
int s, t, st, tt;
};
Fly fly[maxn]; int main(){
ios::sync_with_stdio(false);
while(cin >> n){
mp.cl;
int s, t, last, num;
string city1, city2;
cin >> city1; s = ID(city1);
cin >> city2; t = ID(city2);
cin >> last;
last = last / 100 * 60 + last % 100;
cin >> m;
int S = m<<1, T = m<<1|1;
dinic.init(T + 3);
for(int i = 0; i < m; ++i){
cin >> city1 >> city2 >> num >> fly[i].st >> fly[i].tt;
fly[i].s = ID(city1);
fly[i].t = ID(city2);
fly[i].st = fly[i].st / 100 * 60 + fly[i].st % 100;
fly[i].tt = fly[i].tt / 100 * 60 + fly[i].tt % 100;
if(fly[i].s == s) dinic.addEdge(S, i<<1, INF);
if(fly[i].t == t && fly[i].tt <= last) dinic.addEdge(i<<1|1, T, INF);
dinic.addEdge(i<<1, i<<1|1, num);
for(int j = 0; j < i; ++j){
if(fly[i].t == fly[j].s && fly[i].tt + 30 <= fly[j].st) dinic.addEdge(i<<1|1, j<<1, INF);
else if(fly[j].t == fly[i].s && fly[j].tt + 30 <= fly[i].st) dinic.addEdge(j<<1|1, i<<1, INF);
}
}
cout << dinic.maxflow(S, T) << endl;
}
return 0;
}