给一张n个点m条带权边的有向图，有x个人从起点出发到终点，每个人带的都带相同重量的货物，
规定一条边最多能经过其上权的重量的货物，问最多能带多重的货物？
2 ≤ n ≤ 50, 1 ≤ m ≤ 500, 1 ≤ x ≤ 100 000

二分每个人带的货物
只要把每条边最多能经过几个人算出来，跑流，看最多能让几人通过

考虑点数小，网络流可以乱搞

#include<bits/stdc++.h>
using namespace std;
#define
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#define
                        For(j,m-1) cout<<a[i][j]<<' ';\
                        cout<<a[i][m]<<endl; \
                        } 
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
    int x=0,f=1; char ch=getchar();
    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
    return x*f;
} 
#define
int n,m;
ll x,a[MAXN][MAXN],adj[MAXN][MAXN];
int used[MAXN];
int calc(int s,int t,int c) {
    if (s==t) return 1;
    if (used[s]) return 0;
    used[s]=1;
    For(i,n) if (s!=i&&adj[s][i]>=c){
        if (calc(i,t,c)) {
            adj[s][i]-=c;
            adj[i][s]+=c;
            return 1;
        }
    }
    return 0;
}
int main()
{
//  freopen("CF653D.in","r",stdin);
//  freopen(".out","w",stdout);

    cin>>n>>m>>x;
    MEM(a)
    For(i,m) {
        int x=read(),y=read();
        a[x][y]=read();
    }
    double l=0,r=1e9;
    For(iter,100) {
        double m=(l+r)/2;
        For(i,n) For(j,n) adj[i][j]=(double)a[i][j]/m;
        ll c=100000,ans=0;
        while(c) {
            for(;memset(used,0,sizeof(used)), calc(1,n,c);) ans+=c;
            c/=2;
        }
        if (ans>=x) l=m;else r=m;
    }

    printf("%.8lf\n",r*x);
    return 0;
}