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>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}

using ll=long long;
using R=long double;
const R EPS=1e-9; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r){return(r > EPS)-(r < -EPS);}
inline R sq(R x){return sqrt(max<R>(x,0.0));}

const int dx[8]={1,0,-1,0,1,-1,-1,1};
const int dy[8]={0,1,0,-1,1,1,-1,-1};

// Problem Specific Parameter:
const ll inf=1LL<<50;

// Description: ??°??????????????????(?????????????°??????¨???)
// Verifyed: Many Diffrent Problem 

//Appropriately Changed
using edge=struct{int to,cost,cap,rev;};
using G=vector<vector<edge>>;

//Appropriately Changed
void add_edge(G &graph,int from,int to,int cap,int cost){ 
	graph[from].push_back({to,cost,cap,int(graph[to].size())});
	graph[to].push_back({from,-cost,0,int(graph[from].size())-1});
}

// Description: ??°??????????????????????°??????¨??? 
// TimeComplexity: $ \mathcal{O}(FVE) $ 
// Verifyed: AOJ GRL_6_B

template <typename W> W primal_dual(G &graph,int s,int t,int f,W inf){
	W res=0;
	while(f){
		int n=graph.size(),update;
		vector<W> dist(n,inf);
		vector<int> pv(n,0),pe(n,0);
		dist[s]=0;

		rep(loop,n){
			update=false;
			rep(v,n)rep(i,graph[v].size()){
				edge &e=graph[v][i];
				if(e.cap>0 && chmin(dist[e.to],dist[v]+e.cost)){
					pv[e.to]=v,pe[e.to]=i;
					update=true;
				}
			}
			if(!update) break;
		}

		if(dist[t]==inf) return -1;

		int d=f;
		for(int v=t;v!=s;v=pv[v]) chmin(d,graph[pv[v]][pe[v]].cap);
		f-=d,res+=d*dist[t];

		for(int v=t;v!=s;v=pv[v]){
			edge &e=graph[pv[v]][pe[v]];
			e.cap-=d;
			graph[v][e.rev].cap+=d;
		}
	}
	return res;
}

int main(void){
	int v,e,f;
	cin >> v >> e >> f;
	G graph(v);
	rep(i,e){
		int a,b,c,d;
		cin >> a >> b >> c >> d;
		add_edge(graph,a,b,c,d);
	}

	cout << primal_dual<ll>(graph,0,v-1,f,inf) << endl;
	return 0;
}