description
analysis
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正解二分+DP++DP++DP+判定性问题
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首先可以知道,可能的速度
(距离)一定是某两个点之间的距离 -
O(n2)O(n^2)O(n2)预处理出两点之间的距离,然后排个序,可能的数最多只有n(n+1)2=2001000{n(n+1)\over 2}=20010002n(n+1)=2001000个
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然后二分一个midmidmid,设f[i]f[i]f[i]为第iii个点的最大值,O(n2)O(n^2)O(n2)来转移判定
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最后剩下来的东西分解质因数什么的搞一下就好了
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时间复杂度O(n2log2n22)O(n^2\log_2{n^2\over 2})O(n2log22n2),常数卡的很紧
所以吸臭氧
code
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define MAXN 2005
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)
#define O3 __attribute__((optimize("-O3")))
using namespace std;
double map[MAXN][MAXN];
ll t[MAXN],x[MAXN],y[MAXN],f[MAXN];
ll n,k,tot;
struct node
{
double dis;
ll b,c;
}a[MAXN*MAXN/2];
O3 inline ll read()
{
ll x=0,f=1;char ch=getchar();
while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
O3 inline ll sqr(ll x)
{
return x*x;
}
O3 inline bool cmp(node a,node b)
{
return a.dis<b.dis;
}
O3 inline bool judge(ll mid)
{
double midd=a[mid].dis;
memset(f,0,sizeof(f));
fo(i,1,n)
{
fo(j,0,i-1)if (map[i][j]<=midd)f[i]=max(f[i],f[j]+1);
if (f[i]>=k)return 1;
}
return 0;
}
O3 inline ll gcd(ll x,ll y)
{
return !y?x:gcd(y,x%y);
}
O3 inline void doit(ll x,ll y)
{
ll z=x,b=1;
for (reg i=2;i*i<=x;++i)
{
while (z%(i*i)==0)z/=(i*i),b*=i;
}
ll GCD=gcd(b,y);
printf("%lld %lld %lld\n",b/GCD,z,y/GCD);
}
O3 int main()
{
//freopen("T1.in","r",stdin);
n=read(),k=read();
fo(i,1,n)t[i]=read(),x[i]=read(),y[i]=read();
fo(i,0,n-1)
{
fo(j,i+1,n)
{
a[++tot].b=sqr(x[i]-x[j])+sqr(y[i]-y[j]),a[tot].c=t[j]-t[i];
a[tot].dis=map[i][j]=map[j][i]=1.0*sqrt(a[tot].b)/a[tot].c;
}
}
sort(a+1,a+tot+1,cmp);
ll l=1,r=tot,mid;
while (l<r)
{
mid=(l+r)/2;
judge(mid)?r=mid:l=mid+1;
}
doit(a[r].b,a[r].c);
return 0;
}