題意：給出一張n個點m條邊的圖，從起點1送k個單位的貨物到終點n，其中每條邊有容量限制Ci，且每條邊的花費與貨物經過這條邊的大小成ai 的係數關係，即運輸x的貨物，要ai * x * x的花費，求是否能夠送到終點，且求最小花費。

由於c才為5，很容易想到將其拆邊，假設容量為3，則能依次得到容量大小為1，花費為1， 3， 5的三條邊（1+5+3 = 3*3），然後跑一發費用流即可。

#include <bits/stdc++.h>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <time.h>
#include <vector>
#include <cstdio>
#include <string>
#include <iomanip>
///cout << fixed << setprecision(13) << (double) x << endl;
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#define ls rt << 1
#define rs rt << 1 | 1
#define pi acos(-1.0)
#define eps 1e-8
#define Mp(a, b) make_pair(a, b)
#define asd puts("asdasdasdasdasdf");
typedef long long ll;
typedef pair <int, int> pl;
//typedef __int64 LL;
const int inf = 0x3f3f3f3f;
const int N = 111, M = 500050;

struct node{
    int v, w, c, nxt;    //c--->cap  w--->cost(dis)
}e[M];

int cnt, n, m, k;
int head[N];
int dis[N];    //費用看成距離
int cap[N];        //流量
int cur[N];        //當前弧
int vis[N];
queue <int> q;
int flow, cost;
int st, ed;

void init()
{
    st = 0, ed = n+1;
    cnt = 0;
    memset( head, -1, sizeof( head ) );
}

void add( int u, int v, int c, int w )
{
    e[cnt].v = v;
    e[cnt].c = c;
    e[cnt].w = w;
    e[cnt].nxt = head[u];
    head[u] = cnt++;

    e[cnt].v = u;
    e[cnt].c = 0;
    e[cnt].w = -w;
    e[cnt].nxt = head[v];
    head[v] = cnt++;
}

bool spfa()
{
    memset( dis, inf, sizeof( dis ) );
    memset( vis, 0, sizeof( vis ) );
    while( !q.empty() )    q.pop();
    cap[st] = inf;
    cur[st] = -1;
    dis[st] = 0;
    q.push( st );
    while( !q.empty() ) {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for( int i = head[u]; ~i; i = e[i].nxt ) {
            int v = e[i].v, c = e[i].c, w = e[i].w;
            if( c && dis[v] > dis[u] + w ) {
                dis[v] = dis[u] + w;
                cap[v] = min( c, cap[u] );
                cur[v] = i;
                if( !vis[v] ) {
                    vis[v] = 1;
                    q.push(v);
                }
            }
        }
    }
    if( dis[ed] == inf )
        return 0;
    flow += cap[ed];
    cost += cap[ed] * dis[ed];
    for( int i = cur[ed]; ~i; i = cur[e[i^1].v] ) {
        e[i].c -= cap[ed];
        e[i^1].c += cap[ed];
    }
    return 1;
}

int MCMF()
{
    flow = cost = 0;
    while( spfa() );
    return cost;
}

int main()
{
    while( ~scanf("%d%d%d", &n, &m, &k) ) {
        init();
        for( int i = 1; i <= m; ++i ) {
            int u, v, a, c;
            scanf("%d%d%d%d", &u, &v, &a, &c);
            for( int j = 1; j <= c; ++j ) {
                add( u, v, 1, a * ( 2 * j - 1 ) );
            }
        }
        add( st, 1, k, 0 );
        add( n, ed, inf, 0 );
        int ans = MCMF();
        if( flow < k )
            puts("-1");
        else
            printf("%d\n", ans);
    }
}