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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>

using namespace std;

static const int MAX_V = 110;
static const int INF = (1<<30);

typedef pair<int, int> P;

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

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

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

int min_cost_flow(int s, int t, int f){
	int res = 0;
	fill(h, h + V, 0);
	while (f > 0){
		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;
			for (int i = 0; i < 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(P(dist[e.to], e.to));
				}
			}
		}
		if (dist[t] == INF) return -1;
		for (int v = 0; v < 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;
}

int main(int argc, char const *argv[])
{
	int E, F;
	cin >> V >> E >> F;
	for (int i = 0; i < E; ++i)
	{
		int s, t, c, d;
		cin >> s >> t >> c >> d;
		add_edge(s, t, c, d);
	}
	cout << min_cost_flow(0, V-1, F) << endl;
	return 0;
}