bzoj 2395 Timeismoney —— 最小乘积生成树

题目：https://www.lydsy.com/JudgeOnline/problem.php?id=2395

参考博客：https://www.cnblogs.com/autsky-jadek/p/3959446.html

但复杂度不太会算；

递归边界不要取两个点相等，而是叉积>=0 。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int const xn=,xm=1e4+;
int n,m,fa[xn],dis[xn];
ll ansx,ansy;
struct N{
  int u,v,a,b; ll w;
  bool operator < (const N &y) const
  {return w<y.w;}
}ed[xm];
struct P{
  ll x,y;
  P(ll x=,ll y=):x(x),y(y) {}
};
P operator - (const P &A,const P &B){return P(A.x-B.x,A.y-B.y);}
ll cross(P A,P B){return A.x*B.y-A.y*B.x;}
int rd()
{
  int ret=,f=; char ch=getchar();
  while(ch<''||ch>''){if(ch=='-')f=; ch=getchar();}
  while(ch>=''&&ch<='')ret=ret*+ch-'',ch=getchar();
  return f?ret:-ret;
}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void uni(int x,int y)
{
  if(dis[x]==dis[y])fa[x]=y,dis[y]++;
  else if(dis[x]<dis[y])fa[x]=y;
  else fa[y]=x;
}
P kruskal()
{
  int cnt=; P ret=P(,);
  memset(dis,,sizeof dis);
  for(int i=;i<=n;i++)fa[i]=i;
  sort(ed+,ed+m+);
  for(int i=;i<=m;i++)
    {
      int x=find(ed[i].u),y=find(ed[i].v);
      if(x==y)continue;
      ret.x+=ed[i].a; ret.y+=ed[i].b;
      uni(x,y); cnt++;
      if(cnt==n-)break;
    }
  return ret;
}
bool eql(P A,P B){return A.x==B.x&&A.y==B.y;}
void solve(P A,P B)
{
  for(int i=;i<=m;i++)ed[i].w=ed[i].b*(B.x-A.x)+ed[i].a*(A.y-B.y);
  P C=kruskal();
  if(cross(B-A,C-A)>=)return;//>=0
  //if(eql(A,C)||eql(B,C))return;
  if((C.x*C.y<ansx*ansy)||(C.x*C.y==ansx*ansy&&C.x<ansx))ansx=C.x,ansy=C.y;
  solve(A,C); solve(C,B);
}
int main()
{
  n=rd(); m=rd();
  for(int i=;i<=m;i++)
    ed[i].u=rd()+,ed[i].v=rd()+,ed[i].a=rd(),ed[i].b=rd();
  for(int i=;i<=m;i++)ed[i].w=ed[i].a;
  P A=kruskal();
  for(int i=;i<=m;i++)ed[i].w=ed[i].b;
  P B=kruskal();
  if((A.x*A.y<B.x*B.y)||(A.x*A.y==B.x*B.y&&A.x<B.x))ansx=A.x,ansy=A.y;
  else ansx=B.x,ansy=B.y;
  solve(A,B);
  printf("%lld %lld\n",ansx,ansy);
  return ;
}