正题

bzoj 3754

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题目大意

求最小方差生成树

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解题思路

可以枚举数值之和，然后跑最小生成树

每次加25，这样可以提高速度，且如果平均数接近，那么所选的边和答案会相同，再暴力跑一遍即可

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code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 110
#define M 2021
using namespace std;
int n,m,x,y,mn,mx,top,d[N],fa[N];
double ans;
struct node
{int x,y;double l,L;
}a[M];
int find(int x)
{return fa[x]==x?x:fa[x]=find(fa[x]);
}
bool cmp(node a,node b)
{return a.l<b.l;
}
double min(double a,double b)
{if(a<b)return a;return b;
}
double tr()
{double sum=0,num=0;top=0;for(int i=1;i<=n;++i)fa[i]=i;for(int i=1;i<=m;++i){x=find(a[i].x);y=find(a[i].y);if(x==y)continue;fa[x]=y;d[++top]=i;if(top==n-1)break;}for(int i=1;i<=top;++i)sum+=a[d[i]].L;sum/=top;for(int i=1;i<=top;++i)num+=(a[d[i]].L-sum)*(a[d[i]].L-sum);return num;
}
int read()
{char c=getchar();int ds=0,fs=1;while (c<'0'||c>'9') {if (x=='-') fs=-1;c=getchar();}while (c>='0'&&c<='9') ds=(ds<<3)+(ds<<1)+c-48,c=getchar();return ds*fs;
}
int main()
{n=read();m=read();for(int i=1;i<=m;++i){a[i].x=read();a[i].y=read();a[i].l=a[i].L=read();}ans=1000000000;for(int k=0;k<=10000;k+=25){for(int i=1;i<=m;++i)a[i].l=(a[i].L-1.0*k/(n-1))*(a[i].L-1.0*k/(n-1));sort(a+1,a+1+m,cmp);ans=min(ans,tr());}printf("%.4lf",sqrt(ans/(n-1)));return 0;
}