[国家集训队] 航班安排

对请求拆点建图

对于一个请求，如果 (0) 时刻可以从 (0) 机场到这里，那么 (S) 向它连边，流量 (infty)，费用为 (-w)

结束时间飞回 (0) 小于时间限制，则向着 (T) 连边，费用为 (-w)

两两枚举所有请求，如果来得及就同理连边

最后别忘了限制一下总流量

跑最小费用最大流即可

#include <bits/stdc++.h>
using namespace std;

#define int long long

// Init: init() !!!!!
// Input: make(u,v,cap,cost)
// Solver: solve(s,t)
// Output: ans, cost
namespace flow {
const int N = 100005;
const int M = 1000005;
const int inf = 1e+12;
struct Edge {
    int p, c, w, nxt = -1;
} e[N];
int s, t, tans, ans, cost, ind, bus[N], qhead = 0, qtail = -1, qu[M],vis[N], dist[N];

void graph_link(int p, int q, int c, int w) {
    e[ind].p = q;
    e[ind].c = c;
    e[ind].w = w;
    e[ind].nxt = bus[p];
    bus[p] = ind;
    ++ind;
}
void make(int p, int q, int c, int w) {
    graph_link(p, q, c, w);
    graph_link(q, p, 0, -w);
}
int dinic_spfa() {
    qhead = 0;
    qtail = -1;
    memset(vis, 0x00, sizeof vis);
    memset(dist, 0x3f, sizeof dist);
    vis[s] = 1;
    dist[s] = 0;
    qu[++qtail] = s;
    while (qtail >= qhead) {
        int p = qu[qhead++];
        vis[p] = 0;
        for (int i = bus[p]; i != -1; i = e[i].nxt)
            if (dist[e[i].p] > dist[p] + e[i].w && e[i].c > 0) {
                dist[e[i].p] = dist[p] + e[i].w;
                if (vis[e[i].p] == 0)
                    vis[e[i].p] = 1, qu[++qtail] = e[i].p;
            }
    }
    return dist[t] < inf;
}
int dinic_dfs(int p, int lim) {
    if (p == t)
        return lim;
    vis[p] = 1;
    int ret = 0;
    for (int i = bus[p]; i != -1; i = e[i].nxt) {
        int q = e[i].p;
        if (e[i].c > 0 && dist[q] == dist[p] + e[i].w && vis[q] == 0) {
            int res = dinic_dfs(q, min(lim, e[i].c));
            cost += res * e[i].w;
            e[i].c -= res;
            e[i ^ 1].c += res;
            ret += res;
            lim -= res;
            if (lim == 0)
                break;
        }
    }
    return ret;
}
void solve(int _s,int _t) {
    s=_s; t=_t;
    while (dinic_spfa()) {
        memset(vis, 0x00, sizeof vis);
        ans += dinic_dfs(s, inf);
    }
}
void init() {
    memset(bus, 0xff, sizeof bus);
}
}

struct query {int a,b,s,t,c;} q[205];
int N,M,K,T,t[205][205],f[205][205];

signed main() {
    ios::sync_with_stdio(false);
    cin>>N>>M>>K>>T;
    flow::init();
    for(int i=1;i<=N;i++) {
        for(int j=1;j<=N;j++) {
            cin>>t[i][j];
        }
    }
    for(int i=1;i<=N;i++) {
        for(int j=1;j<=N;j++) {
            cin>>f[i][j];
        }
    }
    for(int i=1;i<=M;i++) {
        cin>>q[i].a>>q[i].b>>q[i].s>>q[i].t>>q[i].c;
        ++q[i].a;
        ++q[i].b;
        flow::make(i,i+M,1,-q[i].c);
        if(q[i].s >= t[1][q[i].a]) {
            flow::make(2*M+1,i,1e9,f[1][q[i].a]);
        }
        if(T-q[i].t >= t[q[i].b][1]) {
            flow::make(i+M,2*M+2,1e9,f[q[i].b][1]);
        }
    }
    for(int i=1;i<=M;i++) {
        for(int j=1;j<=M;j++) {
            if(q[j].s - q[i].t >= t[q[i].b][q[j].a]) {
                flow::make(i+M,j,1e9,f[q[i].b][q[j].a]);
            }
        }
    }
    flow::make(2*M+3,2*M+1,K,0);
    flow::solve(2*M+3,2*M+2);
    cout<<-flow::cost;
}