CF 653D(Delivery Bears-网络流)

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 For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[i]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
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 MAXN (50+10)
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;
}