bzoj2429- 聪明的猴子

题意其实就是说有很多个点，求一组边把它们都连接起来，并且最大的那条边最小。很明显这就是一个最小生成树，是一颗保证最长边最短的树。

代码

刚刚学了个Borůvka算法，于是写了两个。

Borůvka

#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
#include<cmath>
#define Pow(x) ((x)*(x))
using namespace std;
int read() {
	int x=0,f=1;
	char c=getchar();
	for (;!isdigit(c);c=getchar()) if (c=='-') f=-1;
	for (;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int maxh=505;
const int maxn=1e3+10;
const int maxm=maxn*maxn;
int jp[maxh],n,m,all,f[maxn],close[maxn];
struct node {
	double x,y;
} a[maxn];
struct bian {
	int u,v;
	double w;
} e[maxm];
double dist(node &a,node &b) {
	return sqrt(Pow(a.x-b.x)+Pow(a.y-b.y));
}
int find(int x) {
	return f[x]==x?x:f[x]=find(f[x]);
}
double boruvka() {
	for (int i=1;i<=n;++i) f[i]=i;
	e[0].w=1e300;
	double ret=0;
	for (int t=n;t>1;) {
		memset(close,0,sizeof close);
		for (int i=1;i<=all;++i) if (find(e[i].u)!=find(e[i].v)) {
			int fx=find(e[i].u),fy=find(e[i].v);
			if (e[i].w<e[close[fx]].w) close[fx]=i;
			if (e[i].w<e[close[fy]].w) close[fy]=i;
		}
		for (int i=1;i<=n;++i) if (find(i)==i && close[i]) {
			int x=find(e[close[i]].u),y=find(e[close[i]].v);
			if (x!=y) f[x]=y,ret=max(ret,e[close[i]].w),--t;
		}
	}
	return ret;
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("test.in","r",stdin);
	freopen("my.out","w",stdout);
#endif
	m=read();
	for (int i=1;i<=m;++i) jp[i]=read();
	n=read(),all=0;
	for (int i=1;i<=n;++i) a[i].x=read(),a[i].y=read();
	for (int i=1;i<=n;++i) for (int j=i+1;j<=n;++j) e[++all]=(bian){i,j,dist(a[i],a[j])};
	double mst=boruvka();
	int ans=0;
	for (int i=1;i<=m;++i) ans+=(jp[i]>=mst);
	printf("%d\n",ans);
	return 0;
}

Kruskal

#include<cstdio>
#include<cctype>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxh=505;
const int maxn=1e3+10;
const int maxm=maxn*maxn;
struct bian {
	int u,v;
	double w;
	inline bool operator < (const bian &a) const {return w<a.w;}
} e[maxm];
struct P {
	double x,y;
} a[maxn];
double jp[maxn];
int f[maxn];
int find(int x) {
	return f[x]==x?f[x]:f[x]=find(f[x]);
}
double dist(P &a,P &b) {
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int main() {
#ifndef ONLINE_JUDGE
	freopen("test.in","r",stdin);
	freopen("std.out","w",stdout);
#endif
	int n,m;
	scanf("%d",&m);
	for (int i=1;i<=m;++i) scanf("%lf",jp+i);
	scanf("%d",&n);
	for (int i=1;i<=n;++i) scanf("%lf%lf",&a[i].x,&a[i].y);
	int all=0;
	double mst=0;
	for (int i=1;i<=n;++i) for (int j=i+1;j<=n;++j) e[++all]=(bian){i,j,dist(a[i],a[j])};
	sort(e+1,e+all+1);
	for (int i=1;i<=n;++i) f[i]=i;
	for (int i=1,j=0;i<=all && j<n;++i) {
		int u=e[i].u,v=e[i].v,fx=find(u),fy=find(v);
		double w=e[i].w;
		if (fx!=fy) {
			f[fx]=fy;
			mst=max(mst,w);
			++j;
		}
	}
	int ans=0;
	for (int i=1;i<=m;++i) if (jp[i]>=mst) ++ans;
	printf("%d\n",ans);
	return 0;
}