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凸包

参考资料

Andrew 算法

931 Bcpp
#include <bits/stdc++.h>
using namespace std;

const int N=100005;
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y):x(_x),y(_y){}
Point operator-(Point B){return Point(x-B.x,y-B.y);}
}a[N];
double Dis(Point A,Point B){return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}
double Cross(Point A,Point B){return A.x*B.y-A.y*B.x;}
bool cmp(Point A,Point B)
{
return A.x!=B.x?A.x<B.x:A.y<B.y;
}
int s[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i].x>>a[i].y;
}
sort(a+1,a+n+1,cmp);
int m=0;
for(int i=1;i<=n;i++)
{
while(m>1&&Cross(a[s[m]]-a[s[m-1]],a[i]-a[s[m]])<=0)m--;
s[++m]=i;
}
int t=m;
for(int i=n-1;i>=1;i--)
{
while(m>t&&Cross(a[s[m]]-a[s[m-1]],a[i]-a[s[m]])<=0)m--;
s[++m]=i;
}
if(m>1)m--;
double ans=0;
for(int i=1;i<=m;i++)ans+=Dis(a[s[i]],a[s[i%m+1]]);
cout<<fixed<<setprecision(2)<<ans<<'\n';
return 0;
}

Graham 扫描法

911 Bcpp
#include <bits/stdc++.h>
using namespace std;

const int N=100005;
struct Point
{
double x,y;
Point(){}
Point(double _x,double _y):x(_x),y(_y){}
Point operator-(Point B){return Point(x-B.x,y-B.y);}
}a[N];
double Dis(Point A,Point B){return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));}
double Cross(Point A,Point B){return A.x*B.y-A.y*B.x;}
bool cmp(Point A,Point B)
{
double t=Cross(A-a[1],B-a[1]);
return t!=0?t>0:Dis(A,a[1])<Dis(B,a[1]);
}
int s[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i].x>>a[i].y;
if(a[i].y!=a[1].y?a[i].y<a[1].y:a[i].x<a[1].x)swap(a[i],a[1]);
}
sort(a+2,a+n+1,cmp);
int m=0;
for(int i=1;i<=n;i++)
{
while(m>1&&Cross(a[s[m]]-a[s[m-1]],a[i]-a[s[m]])<=0)m--;
s[++m]=i;
}
double ans=0;
for(int i=1;i<=m;i++)ans+=Dis(a[s[i]],a[s[i%m+1]]);
cout<<fixed<<setprecision(2)<<ans<<'\n';
return 0;
}

例题

在城堡外修建一堵围墙,要求围墙离城堡的最近距离不能少于 LL,求出最少需要修建多长的围墙。

给定平面上一些规格相同的信用卡,求其凸包的周长。

给定平面上 nn 个点,求出最短的包围所有点的轮廓且满足任意点到轮廓的距离不小于给定的 LL