洛谷P2604 网络扩容 拆点+费用流

原题链接

这题貌似比较水吧，最简单的拆点，直接上代码了。

#include <bits/stdc++.h>

using namespace std;

#define N 1000
#define M 5000
#define INF 0x3f3f3f3f
#define mp make_pair
#define pii pair<int, int>
#define pb push_back

int n, m, K, S, T;
int d[2*N+5], vis[2*N+5], a[2*N+5], pre[2*N+5];
int t1[M+5], t2[M+5], t3[M+5], t4[M+5];

struct Edge
{
    int from, to, cap, flow, cost;
};
vector<Edge> edges;
vector<int> G[2*N+5];

void addEdge(int u, int v, int cap, int cost)
{
    edges.pb(Edge{u, v, cap, 0, cost}), edges.pb(Edge{v, u, 0, 0, -cost});
    G[u].pb(edges.size()-2), G[v].pb(edges.size()-1);
}

int SPFA(int &flow, int &cost)
{
    memset(d, 0x3f, sizeof d);
    d[S] = 0, vis[S] = 1, pre[S] = 0, a[S] = INF;
    queue<int> q;
    q.push(S);
    while(!q.empty())
    {
        int u = q.front(); q.pop();
        vis[u] = 0;
        for(int i = 0, v; 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;
                pre[e.to] = G[u][i];
                a[e.to] = min(a[u], e.cap-e.flow);
                if(!vis[e.to]) vis[e.to] = 1, q.push(e.to);
            }
        }
    }
    if(d[T] == INF) return 0;
    flow += a[T], cost += d[T]*a[T];
    int u = T;
    while(u != S)
    {
        edges[pre[u]].flow += a[T], edges[pre[u]^1].flow -= a[T];
        u = edges[pre[u]].from;
    }
    return 1;
}

pii minCost()
{
    int flow = 0, cost = 0;
    while(SPFA(flow, cost));
    return mp(flow, cost);
}

void clear()
{
    for(int i = 0; i <= n; ++i) G[i].clear();
    edges.clear();
}

int main()
{
    scanf("%d%d%d", &n, &m, &K);
    for(int i = 1; i <= m; ++i)
    {
        scanf("%d%d%d%d", &t1[i], &t2[i], &t3[i], &t4[i]);
        addEdge(t1[i], t2[i], t3[i], 0);
    }
    S = 0, T = n;
    addEdge(S, 1, INF, 0);
    int ans1;
    printf("%d ", ans1 = minCost().first);
    clear();
    addEdge(S, 1, ans1+K, 0);
    for(int i = 1; i <= n; ++i) addEdge(i, i+n, INF, 0);
    for(int i = 1; i <= m; ++i)
        addEdge(t1[i], t2[i], t3[i], 0), addEdge(t1[i]+n, t2[i], K, t4[i]);
    printf("%d\n", minCost().second);
    return 0;
}