题意

思路

本题做法，给每个月建立两个点 i 和 i'，在建立源点s和汇点t。

①源点s向每个点 i 连一条边，流量为第i个月的产量，花费为生产成本

②每个点 i' 到汇点建一条边，流量为第i个月的销量，花费为（负的！）单位售价

③每个点 i ，向 i' ,(i'+1),(i'+2),(i'+3)...连一条边，流量为INF，花费为储存月数的花费

当s-t的最短路长度大于0时，就停止增广，因为再增广就会使最大利润减小了。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int from, to, cap, flow, cost;       //起点,终点,容量,流量,花费
    Edge(int u, int v, int c, int f, int w):from(u), to(v), cap(c), flow(f), cost(w) {}
};
struct MCMF
{
    int n, m;                //结点数,边数(包括反向弧),源点s,汇点t
    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edges数组中的序号
    bool inq[MAXN];                //是否在队列中
    int d[MAXN];                   //Bellman-Ford
    int p[MAXN];                   //上一条弧
    int a[MAXN];                   //可改进量

    void init(int n)
    {
        this->n = n;
        edges.clear();
        for (int i = 0; i <= n; i++) G[i].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, long long& cost)//SPFA
    {
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(inq, 0, sizeof(inq));
        d[s] = 0; inq[s] = true; p[s] = 0; a[s] = INF;

        queue<int> 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] = true; }
                }
            }
        }
        if (d[t] > 0) return false;
        if (d[t] == INF) return false;
        flow += a[t];
        cost += (long long)d[t] * (long long)a[t];
        for (int u = t; u != s; u = edges[p[u]].from)
        {
            edges[p[u]].flow += a[t];
            edges[p[u]^1].flow -= a[t];
        }
        return true;
    }

    int MinCostMaxFlow(int s, int t, long long& cost)
    {
        int flow = 0; cost = 0;
        while (BellmanFord(s, t, flow, cost));
        return flow;
    }

}solve;

int main()
{
    int T, CASE = 1; scanf("%d", &T);
    while(T--)
    {
        int M, I; scanf("%d%d", &M, &I);
        int S = 0, T = 2*M+1;
        solve.init(T);
        for (int i = 1; i <= M; i++)
        {
            int a, b, c, d, e;
            scanf("%d%d%d%d%d", &a, &b, &c, &d, &e);
            solve.AddEdge(S, i, b, a);
            for (int j = 0; j <= e; j++) if (i+j <= M) solve.AddEdge(i, i+j+M, INF, I*j);
            solve.AddEdge(i+M, T, d, -c);
        }
        long long cost = 0;
        int Max_Flow = solve.MinCostMaxFlow(S, T, cost);
        printf("Case %d: %lld\n", CASE++, -cost);
    }
    return 0;
}

/*
1
2 2
2 10 3 20 2
10 100 7 5 2
*/

UVA ~ 11613 ~ Acme Corporation （流量不固定的有源汇的最小费用流）

题意

思路

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