hoj 2543 (费用流 拆边）

http://acm.hit.edu.cn/hoj/problem/view?id=2543

1.将原图中的每条边(u, v)拆成两条:(u, v, Ci, 0), (u, v, ∞, Ei)

2.购买的每个石头的费用P加一条 (S, 1, inf, P)的边。

3.总的能够花费的费用C可以在我们求最小费用路的时候判断。

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

using namespace std;
typedef long long LL;
const int maxn = 1e3 + 5;
const int maxm = 1e5+ 5;
const int inf = 0x3f3f3f3f;

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;
    }
    void mcmf(int S, int T, int ct)
    {
        sumFlow = 0;
        LL 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,(LL)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;
            if (minCost > ct)
            {
                sumFlow -= ceil((minCost - ct) * 1.0 / dis[T]);
                break;
            }
        }
    }
    void build(int nt, int mt, int ct, int pt)
    {
        init(nt);
        int u, v, c, w;
        addEdge(0, 1, inf, pt);
        while (mt--)
        {
            scanf("%d%d%d%d", &u, &v, &c, &w);
            u++, v++;
            addEdge(u, v, c, 0); addEdge(u, v, inf, w);
            swap(u, v);
            addEdge(u, v, c, 0); addEdge(u, v, inf, w);
        }
    }
    void solve(int nt, int mt, int ct, int pt)
    {
        build(nt, mt, ct, pt);
        mcmf(0, 2, ct);
        printf("%d
", sumFlow);
    }
}my;
int main()
{
    int tcase, n, m, c, p;
    scanf("%d", &tcase);
    while (tcase--)
    {
        scanf("%d%d%d%d",&n, &m, &c, &p);
        my.solve(n, m, c, p);
    }
    return 0;
}