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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/kedebug/p/3043748.html
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