【Codeforces Round #194 (Div. 1)】Codeforces 333E Summer Earnings

Many schoolchildren look for a job for the summer, and one day, when

Gerald was still a schoolboy, he also decided to work in the summer.

But as Gerald was quite an unusual schoolboy, he found quite unusual

work. A certain Company agreed to pay him a certain sum of money if he

draws them three identical circles on a plane. The circles must not

interfere with each other (but they may touch each other). He can

choose the centers of the circles only from the n options granted by

the Company. He is free to choose the radius of the circles himself

(all three radiuses must be equal), but please note that the larger

the radius is, the more he gets paid.

Help Gerald earn as much as possible.

Input The first line contains a single integer n — the number of

centers (3 ≤ n ≤ 3000). The following n lines each contain two

integers xi, yi ( - 104 ≤ xi, yi ≤ 104) — the coordinates of potential

circle centers, provided by the Company.

All given points are distinct.

Output Print a single real number — maximum possible radius of

circles. The answer will be accepted if its relative or absolute error

doesn’t exceed 10 - 6.

从大到小加入边，第一次出现三元环的时候就是最大直径。具体做法用bitset维护和每个点相连的节点。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cmath>
using namespace std;
struct edge
{
int u,v;
double x;
bool operator < (const edge &ee) const
{
return x>ee.x;
}
}a[9000010];
int n,m,x[30010],y[30010];
bitset<30010> b[30010];
double dis(int u,int v)
{
return sqrt((x[u]-x[v])*(x[u]-x[v])+(y[u]-y[v])*(y[u]-y[v]));
}
int main()
{
int i,j,u,v;
scanf("%d",&n);
for (i=0;i<n;i++)
scanf("%d%d",&x[i],&y[i]);
for (i=0;i<n;i++)
for (j=i+1;j<n;j++)
a[++m]=(edge){i,j,dis(i,j)};
sort(a+1,a+m+1);
for (i=1;i<=m;i++)
if ((b[u=a[i].u]&b[v=a[i].v]).any())
{
printf("%.8f\n",a[i].x/2);
return 0;
}
else b[u][v]=b[v][u]=1;
}