题解 小P的生成树

传送门

完全没有思路

• 对复数的和求最值：假设已知最值复数的单位向量，那要构成这个最值向量就要使各复数在该方向向量上的投影之和最大
  对于两个复数，可以求出它们在方向向量上的投影相等时的方向向量
  枚举复数可以得到许多方向向量，这些向量会将单位圆划分成许多区间，这些区间中各复数的投影大小关系不变
  于是可以在每个区间中随便选一个单位向量，取每个向量的投影长正常求最值，最后对每个区间的结果求最值即可
• 对于生成树：生成树的形态只与各边边权的相对大小有关，而与具体权值无关（即只需要能将边排序）

对于这题，对每个区间跑个最大生成树就好

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long
#define reg register int
//#define int long long 

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n, m;
struct edge{int from, to, a, b; double val;}e[N];

namespace force{
	bool vis[15];
	void solve() {
		int lim=1<<m;
		double ans=0;
		for (int s=1,s2,cnt; s<lim; ++s) {
			s2=s; cnt=0;
			ll suma=0, sumb=0;
			double sum=0;
			do {++cnt; s2&=s2-1;} while (s2);
			if (cnt!=n-1) goto jump;
			for (int i=1; i<=n; ++i) vis[i]=0;
			for (int i=0; i<m; ++i) if (s&(1<<i)) {
				vis[e[i+1].from]=vis[e[i+1].to]=1;
				suma+=e[i+1].a, sumb+=e[i+1].b;
			}
			sum=sqrt(suma*suma+sumb*sumb);
			if (sum<ans) goto jump;
			for (int i=1; i<=n; ++i) if (!vis[i]) goto jump;
			// cout<<"sum: "<<sum<<' '<<bitset<10>(s)<<endl;
			ans=sum;
			jump: ;
		}
		printf("%.6lf\n", ans);
		exit(0);
	}
}

namespace task1{
	int fa[N], ans;
	inline int find(int p) {return fa[p]==p?p:fa[p]=find(fa[p]);}
	void solve() {
		int sum=0;
		for (int i=1; i<=n; ++i) fa[i]=i;
		sort(e+1, e+m+1, [](edge a, edge b){return a.a>b.a;});
		int tot=0;
		for (int i=1; i<=m&&tot<n-1; ++i) {
			int f1=find(e[i].from), f2=find(e[i].to);
			if (f1!=f2) {
				fa[f1]=f2;
				sum+=e[i].a;
				++tot;
			}
		}
		ans=abs(sum);
		sum=0;
		for (int i=1; i<=n; ++i) fa[i]=i;
		sort(e+1, e+m+1, [](edge a, edge b){return a.a<b.a;});
		tot=0;
		for (int i=1; i<=m&&tot<n-1; ++i) {
			int f1=find(e[i].from), f2=find(e[i].to);
			if (f1!=f2) {
				fa[f1]=f2;
				sum+=e[i].a;
				++tot;
			}
		}
		ans=max(ans, abs(sum));
		double tem=ans;
		printf("%.6lf\n", tem);
		exit(0);
	}
}

namespace task{
	int fa[N], top;
	const double pi=acos(-1.0);
	double alpha[N];
	inline double calc(double x1, double x2, double y1, double y2) {return (x1-x2)/(y2-y1);}
	inline int find(int p) {return fa[p]==p?p:fa[p]=find(fa[p]);}
	double kruskal(double base) {
		for (int i=1; i<=n; ++i) fa[i]=i;
		for (int i=1; i<=m; ++i) e[i].val=e[i].a*cos(base)+e[i].b*sin(base);
		sort(e+1, e+m+1, [](edge a, edge b){return a.val>b.val;});
		int tot=0; ll suma=0, sumb=0;
		for (int i=1; i<=m&&tot<n-1; ++i) {
			int f1=find(e[i].from), f2=find(e[i].to);
			if (f1!=f2) {
				fa[f1]=f2;
				suma+=e[i].a, sumb+=e[i].b;
				++tot;
			}
		}
		// cout<<"sum: "<<suma<<' '<<sumb<<endl;
		return sqrt(suma*suma+sumb*sumb);
	}
	void solve() {
		alpha[++top]=90; alpha[++top]=270;
		for (int i=1; i<=m; ++i)
			for (int j=i+1; j<=m; ++j) if (e[i].a!=e[j].a && e[i].b!=e[j].b) {
				double t=atan2(e[i].a-e[j].a, e[i].b-e[j].b);
				// cout<<"t: "<<t<<endl;
				alpha[++top]=t; alpha[++top]=(t+pi);
			}
		sort(alpha+1, alpha+top+1);
		double ans=0;
		for (int i=2; i<=top; ++i) ans=max(ans, kruskal((alpha[i-1]+alpha[i])/2));
		ans=max(ans, kruskal(0));
		printf("%.6lf\n", ans);
		exit(0);
	}
}

signed main()
{
	freopen("mst.in", "r", stdin);
	freopen("mst.out", "w", stdout);

	n=read(); m=read();
	bool itgt=1;
	for (int i=1,u,v,a,b; i<=m; ++i) {
		u=read(); v=read(); a=read(); b=read();
		e[i].from=u; e[i].to=v; e[i].a=a; e[i].b=b;
		if (b) itgt=0;
	}
	// if (itgt) task1::solve();
	// else force::solve();
	task::solve();

	return 0;
}