转载:https://blog.csdn.net/acmersch/article/details/5607822
分析:,题目意思时在很多点上取出最短的距离,此题用分治法,先将数据按x排序,在从左到右取出它的最小值,取值时首先是对x排序取出,从左到右依次与中间的值相相减,将小于mmin的点放入数组中,在对该数组进行y排序,进行同样的处理,进行删选,将最后的点都进行处理,对比产生最小值。
代码:
#include<iostream>
include<algorithm>
include<cmath>
#include<stdio.h>
using namespace std;
struct node
{
double x,y;
}f[100010],fr[100010],ff[100010];
int n,nr,nn;
bool cmpx(node a,node b)
{
if (a.x == b.x)
return a.y < b.y;
return a.x < b.x;
}
bool cmpy (node a,node b)
{
if (a.y == b.y)
return a.x < b.x;
return a.y < b.y;
}
double dis(node a,node b)
{
return sqrt( pow( (a.x-b.x),2.0 ) + pow( (a.y-b.y),2.0 ) );
}
double getmin(double a,double b)
{
return a<b?a:b;
}
double findmin(int l,int r)
{
if (l==r)
return 100000000;
else if (l+1 == r)
return dis(f[l],f[r]);
else if (l+2 == r)
return getmin( getmin( dis(f[l],f[r]),dis(f[l],f[l+1]) ),dis(f[r],f[l+1]) );
int mid = (l+r)>>1,i,j;
double mmin = getmin(findmin(l,mid),findmin(mid+1,r)),tmp;
nn = 0;
for (i=l;i<=r;i++)
{
if (fabs(f[i].x - f[mid].x) < mmin)
{
ff[++ nn] = f[i];
}
}
sort(ff+1,ff+nn+1,cmpy);
mid = (1+nn) >> 1;
nr = 0;
for (i=1;i<=nn;i++)
{
if (fabs(ff[i].y - ff[mid].y) < mmin)
{
fr[++ nr] = ff[i];
}
}
for (i=1;i<=nr;i++)
{
for (j=i+1;j<=nr;j++)
{
tmp = dis(fr[i],fr[j]);
mmin = mmin < tmp ? mmin:tmp;
}
}
return mmin;
}
int main ()
{
int i;
while (scanf(“%d”,&n) != EOF && 0 != n)
{
for (i=1;i<=n;i++)
{
scanf(“%lf%lf”,&f[i].x,&f[i].y);
}
sort(f+1,f+n+1,cmpx);
printf(“%.2f/n”,findmin(1,n)/2.0);
}
return 0;
}
分析:,题目意思时在很多点上取出最短的距离,此题用分治法,先将数据按x排序,在从左到右取出它的最小值,取值时首先是对x排序取出,从左到右依次与中间的值相相减,将小于mmin的点放入数组中,在对该数组进行y排序,进行同样的处理,进行删选,将最后的点都进行处理,对比产生最小值。
代码:
#include<iostream>