G. 铁路修复计划
二分答案，改变边的权值，找最小生成树即可。
类似的思想还可以用在单度限制最小生成树和最优比例生成树上。
#include<iostream>
#include<cmath>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<stack>
#include<cstdio>
#include<cstdlib>
using namespace std;
typedef long long int LL;
double k;
int n,m;double M;
struct edge
{
int from;int to;
LL cost;int foreign;
bool operator <(const edge &b)const
{
return (cost+foreign*(k-1.)*cost)<(b.cost+b.foreign*(k-1.)*b.cost);
}
};
const int maxn=100000+5;
edge e[maxn];
int belong[maxn];
int be(int x)
{
if(x==belong[x])return x;
else return belong[x]=be(belong[x]);
}
bool ex(double mid)
{
k=mid;
sort(e,e+m);
for(int i=1;i<=n;i++)belong[i]=i;
double ansnow=0;
for(int i=0;i<m&&ansnow<=M;i++)
{
int u=e[i].from,v=e[i].to;
u=be(u);v=be(v);
//cout<<"ll"<<u<<v<<endl;
if(u!=v){ansnow+=(e[i].cost+e[i].foreign*(k-1.)*e[i].cost),belong[v]=u;}
//cout<<ansnow<<endl;
}
return ansnow<=M;
}
int main()
{
// freopen("t.txt","r",stdin);
LL MM;
scanf("%d%d%lld",&n,&m,&MM);
M=(double)MM;
//cout<<M<<endl;
for(int i=0;i<m;i++)
scanf("%d%d%d%d",&e[i].from,&e[i].to,&e[i].cost,&e[i].foreign);
double l=1.0,r=1e+15;
while((r-l)>1e-8)
{
//cout<<"double"<<l<<" "<<r<<endl;
double mid=(r+l)/2.;
if(ex(mid))l=mid;
else r=mid;
}
printf("%.6lf",(l+r)/2.);
return 0;
}