题目:
给出若干个点
求三个点构成的周长最小的三角形的周长(我们认为共线的三点也算三角形)
题解:
可以参考平面最近点对的做法
只不过合并的时候改成枚举三个点更新周长最小值,其他的和最近点对大同小异
2#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 200010
#define INF 1e20
using namespace std;
int n;
struct point
{
double x,y;
point () {};
point (double _x,double _y)
{
x=_x,y=_y;
}
point operator - (const point &a)const
{
return point(x-a.x,y-a.y);
}
bool operator < (const point &a)const
{
return x<a.x;
}
double norm()
{
return sqrt(x*x+y*y);
}
}p[N];
double calc(const point &x,const point &y,const point &z)
{
return (x-y).norm()+(y-z).norm()+(x-z).norm();
}
double solve(int l,int r)
{
if (r==l) return INF;
int mid=l+r>>1;
double xmid=(p[mid].x+p[mid+1].x)/2;
double ret=min(solve(l,mid),solve(mid+1,r));
static point a[N],b[N],c[N];
int pos=l,i=l,j=mid+1,b_n=0,c_n=0;
while (pos<=r)
{
if (i<=mid && (p[i].y<p[j].y || j>r))
{
if (p[i].x+ret/2>xmid)
b[++b_n]=p[i];
a[pos++]=p[i++];
}
else
{
if (p[j].x-ret/2<xmid)
c[++c_n]=p[j];
a[pos++]=p[j++];
}
}
for (i=l;i<=r;i++)
p[i]=a[i];
if (r-l<2) return INF;
/*
for (int i=1;i<=b_n;i++)
for (int j=i+1;j<=b_n;j++)
for (int k=1;k<=c_n;k++)
if (i!=j) ret=min(ret,(b[i]-b[j]).norm()+(b[i]-c[k]).norm()+(b[j]-c[k]).norm());
for (int i=1;i<=b_n;i++)
for (int j=1;j<=c_n;j++)
for (int k=j+1;k<=c_n;k++)
if (j!=k) ret=min(ret,(b[i]-c[j]).norm()+(b[i]-c[k]).norm()+(c[j]-c[k]).norm());
*/
// /*
for (int i=1,j=1;i<=b_n;i++)
{
while (j<=c_n && b[i].y-c[j].y>ret/2) j++;
for (int k=j;k<=c_n && abs(b[i].y-c[k].y)<ret/2;k++)
for (int h=k+1;h<=c_n && abs(b[i].y-c[h].y)<ret/2;h++)
ret=min(ret,calc(b[i],c[k],c[h]));
}
for (int i=1,j=1;i<=c_n;i++)
{
while (j<=b_n && c[i].y-b[j].y>ret/2) j++;
for (int k=j;k<=b_n && abs(c[i].y-b[k].y)<ret/2;k++)
for (int h=k+1;h<=b_n && abs(c[i].y-b[h].y)<ret/2;h++)
ret=min(ret,calc(c[i],b[k],b[h]));
}
// */
return ret;
}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
scanf("%lf%lf",&p[i].x,&p[i].y);
sort(p+1,p+1+n);
printf("%.6lf",solve(1,n));
return 0;
}