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 <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i=0;i<(n);i++)
const int INF = 1e9;

struct edge { int to, cap, cost, rev; };
int V;
const int MAX_V = 100;
vector<edge> G[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].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;
    while (f > 0) {
        fill(dist, dist + V, INF);
        dist[s] = 0;
        bool update = true;
        while (update) {
            update = false;
            for (int v = 0; v < V; v++) {
                if (dist[v] == INF) 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) {
                        dist[e.to] = dist[v] + e.cost;
                        prevv[e.to] = v;
                        preve[e.to] = i;
                        update = true;
                    }
                }
            }
        }
 
        if (dist[t] == INF) {
            return -1;
        }
 
        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 * dist[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(void){
	cin >> V;
	int e, f; cin >> e >> f;
	rep(i, e){
		int u, v, c, d;
		cin >> u >> v >> c >> d;
		add_edge(u, v, c, d);
	}
	printf("%d\n", min_cost_flow(0, V - 1, f));
}