题目链接：点击打开链接
思路：
1看这些束缚条件和数据量， 我们就想到了网络流。
由于有时间， 我们可以构造2元组(u, t)表示在城市u，时间t这个状态， 以这样的2元组作为结点跑最大流， 惋惜这样结点高达150*24*60， 会TLE。
我们可以斟酌枚举任意两个航线， 如果满足关系， 就建边， 跑最大流。
斟酌到每一个航线只能经过1次， 我们把航线当作结点， 拆点跑最大流便可。
细节参见代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef long double ld;
const double eps = 1e⑹;
const double PI = acos(⑴);
const int mod = 1000000000 + 7;
const int INF = 0x3f3f3f3f;
const int seed = 131;
const ll INF64 = ll(1e18);
const int maxn = 10000 + 10;
int T,n,m;
struct Edge {
int from, to, cap, flow;
};
bool operator < (const Edge& a, const Edge& b) {
return a.from < b.from || (a.from == b.from && a.to < b.to);
}
struct Dinic {
int n, m, s, t; // 结点数， 边数（包括反向弧）， 源点编号， 汇点编号
vector<Edge> edges; // 边表， edges[e]和edges[e^1]互为反向弧
vector<int> G[maxn]; // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
bool vis[maxn]; // BFS使用
int d[maxn]; // 从出发点到i的距离
int cur[maxn]; // 当前弧指针
void init(int 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});
m = edges.size();
G[from].push_back(m⑵);
G[to].push_back(m⑴);
}
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(s);
vis[s] = 1;
d[s] = 0;
while(!Q.empty()) {
int x = Q.front(); Q.pop();
for(int i = 0; i < G[x].size(); i++) {
Edge& e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow) { //只斟酌残量网络中的弧
vis[e.to] = 1;
d[e.to] = d[x] + 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++) { //上次斟酌的弧
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;
}
}g;
int u, v, c, kase = 0, t, num[maxn], t1[maxn], t2[maxn], a[maxn], b[maxn];
char s1[11], s2[11], s3[maxn][10], s4[maxn][10];
map<string, int> p;
int main() {
while(~scanf("%d", &n)) {
scanf("%s", s1); p.clear();
scanf("%s", s2); kase = 0;
p[s1] = ++kase; if(!p.count(s2)) p[s2] = ++kase;
int start = p[s1], endd = p[s2];
scanf("%s%d", s1, &m);
t = ((s1[0]-'0')*10+s1[1]-'0')*60+(s1[2]-'0')*10+s1[3]-'0';
for(int i = 1; i <= m; i++) {
scanf("%s%s%d%s%s", s3[i], s4[i], &num[i], s1, s2);
t1[i] = ((s1[0]-'0')*10+s1[1]-'0')*60+(s1[2]-'0')*10+s1[3]-'0';
t2[i] = ((s2[0]-'0')*10+s2[1]-'0')*60+(s2[2]-'0')*10+s2[3]-'0';
if(!p.count(s3[i])) p[s3[i]] = ++kase;
if(!p.count(s4[i])) p[s4[i]] = ++kase;
a[i] = p[s3[i]]; b[i] = p[s4[i]];
}
g.init(2*m + 5);
int res = m;
int src = 2*m + 1, stc = 2*m + 2;
for(int i = 1; i <= m; i++) {
int id1 = b[i];
int cur = a[i];
g.AddEdge(i, i+m, num[i]);
if(start == cur) g.AddEdge(src, i, INF);
if(id1 == endd && t2[i] <= t) g.AddEdge(m+i, stc, INF);
for(int j = 1; j <= m; j++) {
if(i == j) continue;
int id2 = a[j];
if(id1 != id2) continue;
if(t2[i] + 30 <= t1[j]) g.AddEdge(i+m, j, INF);
}
}
printf("%d\n", g.Maxflow(src, stc));
}
return 0;
}