HDU 6437 Problem L.Videos——费用流

s到每个人容量为1，费用为0的边

每个人到每个视频建容量为1， 费用为0的边

每个视频拆点，建容量为1，费用为-w的边

视频之间若时间不覆盖，则容量为1的边，若两视频op相同，费用为W， 否则为0

每个视频到t建容量为1， 费用为0的边

跑一遍s-t最小费用流，答案取负即可

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 1000;
const int INF = 0x3f3f3f3f;
struct Edge {
    int from, to, cap, flow, cost;
};
struct MCMF {
    int n, m, s, t;
    vector edges;
    vector G[maxn];
    int inq[maxn];
    int d[maxn];
    int p[maxn];
    int a[maxn];

    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, int cost) {
        edges.push_back((Edge){from, to, cap, 0, cost});
        edges.push_back((Edge){to, from, 0, 0, -cost});
        m = edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BellmanFord(int s, int t, int &flow, int &cost) {
        for (int i = 0; i < n; i++) d[i] = INF;
        memset(inq, 0, sizeof(inq));
        d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;

        queue q;
        q.push(s);
        while (!q.empty()) {
            int u = q.front(); q.pop();
            inq[u] = 0;
            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if (!inq[e.to]) { q.push(e.to); inq[e.to] = 1; }
                }
            }
        }
        if (d[t] == INF) return false;
        flow += a[t];
        cost += d[t] * a[t];
        int u = t;
        while (u != s) {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }
    int mincost(int s, int t) {
        int flow = 0, cost = 0;
        while (BellmanFord(s, t, flow, cost));
        return cost;
    }
}mcmf;
struct Data {
    int s, t, w, op;
    bool operator < (const Data &atr) const {
        return s < atr.s;
    }
}data[maxn];
int main() {
    int T, n, m, k, W;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d%d", &n, &m, &k, &W);
        mcmf.init(k+2*m+5);
        int s = 0, t = k + 2*m + 1;
        for (int i = 1; i <= k; i++) mcmf.addedge(s, i, 1, 0);
        for (int i = 0; i < m; i++) {
            scanf("%d%d%d%d", &data[i].s, &data[i].t, &data[i].w, &data[i].op);
        }
        sort(data, data + m);
        for (int i = 1; i <= k; i++) {
            for (int j = 0; j < m; j++) {
                mcmf.addedge(i, k+2*j+1, 1, 0);
            }
        }
        for (int i = 0; i < m; i++) mcmf.addedge(k+i*2+1, k+i*2+2, 1, -data[i].w);
        for (int i = 0; i < m; i++) {
            for (int j = i + 1; j < m; j++) {
                if (data[i].t <= data[j].s || data[i].s >= data[j].t) {
                    if (data[i].op == data[j].op) {
                        mcmf.addedge(k+i*2+2, k+j*2+1, 1, W);
                    }
                    else {
                        mcmf.addedge(k+i*2+2, k+j*2+1, 1, 0);
                    }
                }
            }
        }
        for (int i = 0; i < m; i++) {
            mcmf.addedge(k+i*2+2, t, 1, 0);
        }
        printf("%d\n", -mcmf.mincost(s, t));
    }
    return 0;
}