[BZOJ3080]Minimum Variance Spanning Tree/[BZOJ3754]Tree之最小方差树

[BZOJ3080]Minimum Variance Spanning Tree/[BZOJ3754]Tree之最小方差树

题目大意:

给定一个\(n(n\le50)\)个点,\(m(m\le1000)\)条边的带权无向图,每条边的边权为\(w_i(w_i\le50)\)。求最小方差生成树。

3080数据范围:\(n\le50,m\le1000,w_i\le50\)
3754数据范围:\(n\le100,m\le1000,w_i\le100\)

其中3754询问的是最小标准差。

思路:

由于\(w_i\)很小,因此我们可以枚举树上的边权和\(\sum w_i\),以\((w_i-\bar w)^2\)为新的边权做最小生成树。若最后树上的\(\sum w_i=\)一开始枚举的值,那么就更新答案。

源代码(3080):

#include<cstdio>
#include<cctype>
#include<algorithm>
inline int getint() {
	register char ch;
	while(!isdigit(ch=getchar()));
	register int x=ch^'0';
	while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
	return x;
}
const int N=51,M=1001;
double d;
inline double sqr(const double &x) {
	return x*x;
}
struct Edge {
	int u,v,w;
	bool operator < (const Edge &rhs) const {
		return sqr(w-d)<sqr(rhs.w-d);
	}
};
Edge edge[M];
class DisjointSet {
	private:
		int anc[N];
		int find(const int &x) {
			return x==anc[x]?x:anc[x]=find(anc[x]);
		}
	public:
		void reset(const int &n) {
			for(register int i=1;i<=n;i++) anc[i]=i;
		}
		void merge(const int &x,const int &y) {
			anc[find(x)]=find(y);
		}
		bool same(const int &x,const int &y) {
			return find(x)==find(y);
		}
};
DisjointSet djs;
int main() {
	for(register int i=1;;i++) {
		const int n=getint(),m=getint();
		if(n==0&&m==0) return 0;
		for(register int i=1;i<=m;i++) {
			edge[i].u=getint();
			edge[i].v=getint();
			edge[i].w=getint();
		}
		d=0;
		std::sort(&edge[1],&edge[m]+1);
		int l=0,r=0;
		for(register int i=1;i<n;i++) l+=edge[i].w;
		for(register int i=m;i>m-n+1;i--) r+=edge[i].w;
		double ans=1e18;
		for(register int i=l;i<=r;i++) {
			d=1.*i/(n-1);
			std::sort(&edge[1],&edge[m]+1);
			djs.reset(n);
			int sum1=0;
			double sum2=0;
			for(register int i=1;i<=m;i++) {
				const int &u=edge[i].u,&v=edge[i].v;
				if(djs.same(u,v)) continue;
				djs.merge(u,v);
				sum1+=edge[i].w;
				sum2+=sqr(edge[i].w-d);
			}
			if(sum1==i) {
				ans=std::min(ans,sum2/(n-1));
			}
		}
		printf("Case %d: %.2f\n",i,ans);
	}
}

源代码(3754):

#include<cmath>
#include<cstdio>
#include<cctype>
#include<algorithm>
inline int getint() {
	register char ch;
	while(!isdigit(ch=getchar()));
	register int x=ch^'0';
	while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
	return x;
}
const int N=101,M=2001;
double d;
inline double sqr(const double &x) {
	return x*x;
}
struct Edge {
	int u,v,w;
	bool operator < (const Edge &rhs) const {
		return sqr(w-d)<sqr(rhs.w-d);
	}
};
Edge edge[M];
class DisjointSet {
	private:
		int anc[N];
		int find(const int &x) {
			return x==anc[x]?x:anc[x]=find(anc[x]);
		}
	public:
		void reset(const int &n) {
			for(register int i=1;i<=n;i++) anc[i]=i;
		}
		void merge(const int &x,const int &y) {
			anc[find(x)]=find(y);
		}
		bool same(const int &x,const int &y) {
			return find(x)==find(y);
		}
};
DisjointSet djs;
int main() {
	const int n=getint(),m=getint();
	for(register int i=1;i<=m;i++) {
		edge[i].u=getint();
		edge[i].v=getint();
		edge[i].w=getint();
	}
	std::sort(&edge[1],&edge[m]+1);
	int l=0,r=0;
	for(register int i=1;i<n;i++) l+=edge[i].w;
	for(register int i=m;i>m-n+1;i--) r+=edge[i].w;
	double ans=1e18;
	for(register int i=l;i<=r;i++) {
		d=1.*i/(n-1);
		std::sort(&edge[1],&edge[m]+1);
		djs.reset(n);
		int sum1=0;
		double sum2=0;
		for(register int i=1;i<=m;i++) {
			const int &u=edge[i].u,&v=edge[i].v;
			if(djs.same(u,v)) continue;
			djs.merge(u,v);
			sum1+=edge[i].w;
			sum2+=sqr(edge[i].w-d);
		}
		if(sum1==i) {
			ans=std::min(ans,sum2/(n-1));
		}
	}
	printf("%.4f\n",sqrt(ans));
}
posted @ 2018-09-30 14:12  skylee03  阅读(226)  评论(0编辑  收藏  举报