网络流24题（10）餐巾计划问题（最小费用最大流）

思路：

1. 把每天分成两个集：Xi（第 i 天用剩的餐巾），Yi（第 i 天需要的餐巾）

2. s 向 X 引弧，容量为 ri，费用为 0；Y 向 t 引弧，容量为 ri，费用为 0；s 向 Y 引弧，容量为无穷大，费用为 p

3. 由于每天用剩下的餐巾可以分为下面三种情况：

   a. 什么也不做，留着当做第二天的旧餐巾： Xi 向 Xi+1 引弧，容量为无穷大，费用为 0;

   b. 送给快洗部，相当于：Xi 向 Yi+m 引弧，容量为无穷大，费用为 f;

   c. 送给慢洗部，相当于：Xi 向 Yi+n 引弧，容量为无穷大，费用为 s;

4. 求上面二分图的最小费用最大流，因为最大流一定能保证 Yi->t 为满弧，满足约束条件。最小的费用即是要输出的结果。

#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

const int MAXN = 2010;
const int INFS = 0x3FFFFFFF;

struct edge {
    int from, to, cap, flow, cost;
    edge(int _from, int _to, int _cap, int _flow, int _cost)
        : from(_from), to(_to), cap(_cap), flow(_flow), cost(_cost) {}
};

class MCMF {
public:
    void initdata(int n) {
        this->n = n;
        edges.clear();
        for (int i = 0; i < n; i++)
            G[i].clear();
    }
    void addedge(int u, int v, int cap, int cost) {
        edges.push_back(edge(u, v, cap, 0, cost));
        edges.push_back(edge(v, u, 0, 0, -cost));
        G[u].push_back(edges.size() - 2);
        G[v].push_back(edges.size() - 1);
    }

    bool SPFA(int s, int t, int& flow, int& cost) {
        for (int i = 0; i < n; i++)
            d[i] = INFS, inq[i] = false;
        queue<int> Q;
        Q.push(s);
        d[s] = 0, inq[s] = true, p[s] = 0, a[s] = INFS;
        while (!Q.empty()) {
            int u = Q.front(); Q.pop();
            inq[u] = false;
            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]) {
                        inq[e.to] = true;
                        Q.push(e.to);
                    }
                }
            }
        }
        if (d[t] == INFS) return false;
        flow += a[t];
        cost += a[t] * d[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 (SPFA(s, t, flow, cost));
        return cost;
    }
private:
    vector<edge> edges;
    vector<int> G[MAXN];
    int n, d[MAXN], p[MAXN], a[MAXN];
    bool inq[MAXN];
};

MCMF mcmf;

int main() {
    int n, cost;
    int fast, fastcost, slow, slowcost;
    scanf("%d%d", &n, &cost);
    scanf("%d%d", &fast, &fastcost);
    scanf("%d%d", &slow, &slowcost);
    int s = 0, t = 2*n+1;
    mcmf.initdata(t + 1);
    for (int i = 1; i <= n; i++) {
        int need;
        scanf("%d", &need);
        mcmf.addedge(s, i, need, 0);
        mcmf.addedge(i+n, t, need, 0);
        mcmf.addedge(s, i+n, INFS, cost);
    }
    for (int i = 1; i <= n; i++) {
        if (i + 1 <= n)
            mcmf.addedge(i, i+1, INFS, 0);
        if (i + fast <= n)
            mcmf.addedge(i, i+n+fast, INFS, fastcost);
        if (i + slow <= n)
            mcmf.addedge(i, i+n+slow, INFS, slowcost);
    }
    printf("%d\n", mcmf.mincost(s, t));
    return 0;
}