Problem: Examples

Input

4 5 2
0 1 2 1
0 2 1 2
1 2 1 1
1 3 1 3
2 3 2 1


Output

6


Input




Output

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n)-1;i>=0;i--)
#define rrep2(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define all(a) (a).begin(),(a).end()

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> vi;
typedef vector<P> vp;
typedef vector<ll> vll;

const int INF = 99999999;
const int MAX_V = 100000;

struct edge { int to, cap, cost, rev; };

int V, E, F;
vector<edge> G[MAX_V];
int h[MAX_V];                   // potential
int dist[MAX_V];
int prevv[MAX_V], preve[MAX_V]; // privious vertex, edge

void add_edge(int from, int to, int cap, int cost) {
    G[from].emplace_back((edge){to, cap, cost, (int)G[to].size()});
    G[to].emplace_back((edge){from, 0, -cost, (int)G[from].size() - 1});
}

// get min cost flow from s to t
// if we can flow f, then return -1
int min_cost_flow(int s, int t, int f) {
    int res = 0;
    fill(h, h + V, 0);
    while (f > 0) {
        // update h by dijkstra
        priority_queue<P, vector<P>, greater<P>> que;
        fill(dist, dist + V, INF);
        dist[s] = 0;
        que.push(P(0, s));
        while (!que.empty()) {
            P p = que.top(); que.pop();
            int v = p.second;
            if (dist[v] < p.first) continue;
            rep(i, G[v].size()) {
                edge &e = G[v][i];
                if (e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
                    dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
                    prevv[e.to] = v;
                    preve[e.to] = i;
                    que.push(P(dist[e.to], e.to));
                }
            }
        }
        if (dist[t] == INF) {
            // no more flow
            return -1;
        }
        rep(v, V) h[v] += dist[v];

        // flow as much as possible along the minimum path from s to t
        int d = f;
        for (int v = t; v != s; v = prevv[v]) {
            d = min(d, (int)G[prevv[v]][preve[v]].cap);
        }
        f -= d;
        res += d * h[t];
        for (int v = t; v != s; v = prevv[v]) {
            edge &e = G[prevv[v]][preve[v]];
            e.cap -= d;
            G[v][e.rev].cap += d;
        }
    }
    return res;
}

signed main() {
    cin >> V >> E >> F;
    rep(i, E) {
        int u, v, c, d;
        cin >> u >> v >> c >> d;
        add_edge(u, v, c, d);
    }

    cout << min_cost_flow(0, V - 1, F) << endl;
}