UVA - 1161 Objective: Berlin(最大流＋时序模型)

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，现在问一天内最多有多少人能从A飞到B，可以在其他城市中转

解题思路：将飞机票拆点，拆成i–>i ＋ m,容量为座位数。
接着判断一下，航线之间的连线
如果航线的起点是A的话，那么就和超级源点相连，容量为INF
如果航线的终点是B且到达时间小于等于最晚时间，那么连线，容量为INF
如果航线i的终点和航线j的起点相同，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 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<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 160int 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\n");        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\n", ans);}int main() {    while (scanf("%d\n", &n) != EOF)  solve();    return 0;}