hdu 3667 (拆边 mcmf)

注意题目中 边的容量 <= 5.可以把费用权值 a *f ^2化归成 a * f2, 即第一条边费用为 1 * a, 第二条 为 (4 - 1) * a, 第三条为 (9  - 4) * a。。。。。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

const int maxn = 110;
const int maxm = 25000;
const int inf = 0x3f3f3f3f;
const int ww[5] = {1, 3, 5, 7, 9};

struct MCMF
{
    struct Edge
    {
        int v, c, w, next;
    }p[maxm << 1];
    int e, head[maxn], dis[maxn], pre[maxn], cnt[maxn], sumFlow, n;
    bool vis[maxn];
    void init(int nt)
    {
        e = 0; n = nt;
        memset(head, -1, sizeof(head[0]) * (n + 2));
    }
    void addEdge(int u, int v, int c, int w)
    {
        p[e].v = v; p[e].c = c; p[e].w = w; p[e].next = head[u]; head[u] = e++;
        swap(u, v);
        p[e].v = v; p[e].c = 0; p[e].w =-w; p[e].next = head[u]; head[u] = e++;
    }
    bool spfa(int S, int T)
    {
        queue <int> q;
        for (int i = 0; i <= n; ++i)
            vis[i] = cnt[i] = 0, pre[i] = -1, dis[i] = inf;
        vis[S] = 1; dis[S] = 0;
        q.push(S);
        while (!q.empty())
        {
            int u = q.front(); q.pop();
            vis[u] = 0;
            for (int i = head[u]; i + 1; i = p[i].next)
            {
                int v = p[i].v;
                if (p[i].c && dis[v] > dis[u] + p[i].w)
                {
                    dis[v] = dis[u] + p[i].w;
                    pre[v] = i;
                    if (!vis[v])
                    {
                        q.push(v);
                        vis[v] = 1;
                        if (++cnt[v] > n) return 0;
                    }
                }
            }
        }
        return dis[T] != inf;
    }
    int mcmf(int S, int T)
    {
        sumFlow = 0;
        int minFlow = 0, minCost = 0;
        while (spfa(S, T))
        {
            minFlow = inf + 1;
            for (int i = pre[T]; i + 1; i = pre[ p[i ^ 1].v ])
                minFlow = min(minFlow, p[i].c);
            sumFlow += minFlow;
            for (int i = pre[T]; i + 1; i = pre[ p[i ^ 1].v ])
                p[i].c -= minFlow, p[i ^ 1].c += minFlow;
            minCost += dis[T] * minFlow;
        }
        return minCost;
    }
    void build(int nt, int mt, int kt)
    {
        init(nt);
        addEdge(0, 1, kt, 0);
        int u, v, c, w;
        for (int i = 0; i < mt; ++i)
        {
            scanf("%d%d%d%d", &u, &v, &w, &c);
            for (int j = 0; j < c; ++j)
                addEdge(u, v, 1, w * ww[j]);
        }
    }
    void solve(int nt, int mt, int kt)
    {
        build(nt, mt, kt);
        int ans = mcmf(0, n);
    //    cout <<"sumFlow = "  << sumFlow << " cost = " << ans << endl;
        if (sumFlow != kt)
            printf("-1
");
        else
            printf("%d
", ans);
    }
}my;

int main()
{
    int n, m, k;
    while (~scanf("%d%d%d", &n, &m, &k))
        my.solve(n, m, k);
    return 0;
}