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:
//#define __USE_MINGW_ANSI_STDIO 0
#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
const int INF = (1LL<<30);
const ll LLINF = (1LL<<60);
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;
//#define int ll

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

#define MAX_V 10000
struct edge { int to, cap, cost, rev; };

int V;
vector<edge> G[MAX_V];
int h[MAX_V];
int dist[MAX_V];
int prevv[MAX_V], preve[MAX_V];

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

// s??????t????????????f???????°??????¨???
int min_cost_flow(int s, int t, int f) {
  int res = 0;
  fill(h, h+V, 0);
  while(f > 0) {
    priority_queue<PII, vector<PII>, greater<PII>> que;
    fill(dist, dist+V, INF);
    dist[s] = 0;
    que.push({0, s});
    while(que.size()) {
      PII p = que.top(); que.pop();
      int v = p.second;
      if(dist[v] < p.first) continue;
      for(int i=0; i<(int)G[v].size(); ++i) {
        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({dist[e.to], e.to});
        }
      }
    }
    // ????????\???????????????
    if(dist[t] == INF) return -1;
    REP(v, V) h[v] += dist[v];

    int d = f;
    for(int v = t; v != s; v = prevv[v]) {
      d = min(d, 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(void)
{
  int e, f;
  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;

  return 0;
}