HIT暑期集训 计算几何基础二

凸包，Graham扫描法模板（以洛谷P2742为例），网上的一个凸包详解相关博客：https://blog.csdn.net/ZCY19990813/article/details/98034495

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100005
using namespace std;
double X,Y;
struct node
{
    double x,y;
}e[maxn],stk[maxn];
bool cmp(node a,node b)
{
    if (a.y==b.y) return a.x<b.x;
    return a.y<b.y;
}
bool cmp2(node a,node b)//极角排序 
{
    if (atan2(a.y-Y,a.x-X)==atan2(b.y-Y,b.x-X)) return a.x<b.x;
    return (atan2(a.y-Y,a.x-X))<(atan2(b.y-Y,b.x-X));
}
double cross(node a,node b,node c)
{
    node ab,ac;
    ab.x=b.x-a.x;
    ab.y=b.y-a.y;
    ac.x=c.x-a.x;
    ac.y=c.y-a.y;
    return ab.x*ac.y-ab.y*ac.x;
}
double getdis(node a,node b)
{
    node ab;
    ab.x=b.x-a.x;
    ab.y=b.y-a.y;
    return sqrt(ab.x*ab.x+ab.y*ab.y);
}
int main()
{
    int i,n,top;
    double ans;
    scanf("%d",&n);
    for (i=0;i<n;i++) scanf("%lf%lf",&e[i].x,&e[i].y);
    if (n<3) 
    {
        if (n==1) printf("0
");
        else if (n==2) printf("%.2lf
",getdis(e[0],e[1]));
        return 0;
    }
    sort(e,e+n,cmp);
    top=-1;
    stk[++top]=e[0];
    X=e[0].x;Y=e[0].y;
    sort(e+1,e+n,cmp2);
    stk[++top]=e[1];
    for (i=2;i<n;i++)
    {
        while (cross(stk[top-1],stk[top],e[i])<0) top--;
        stk[++top]=e[i];
    }
    ans=0;
    for (i=1;i<=top;i++)
    {
        ans+=getdis(stk[i],stk[i-1]);
    } 
    ans+=getdis(stk[0],stk[top]);
    printf("%.2lf
",ans); 
    return 0;
} 

凸包+旋转卡壳模板题（洛谷P1452）

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100005
#define eps 1e-8
using namespace std;
double X,Y;
int n,top;
struct node
{
    double x,y;
}e[maxn],stk[maxn];
double cross(node a,node b,node c)
{
    node ab,ac;
    ab.x=b.x-a.x;
    ab.y=b.y-a.y;
    ac.x=c.x-a.x;
    ac.y=c.y-a.y;
    return ab.x*ac.y-ab.y*ac.x;
}
double getdis(node a,node b)
{
    node ab;
    ab.x=b.x-a.x;
    ab.y=b.y-a.y;
    return ab.x*ab.x+ab.y*ab.y;
}
bool cmp(node a,node b)
{
    if (a.y==b.y) return a.x<b.x;
    return a.y<b.y;
}
bool cmp2(node a,node b)//极角排序 
{
    if (atan2(a.y-Y,a.x-X)==atan2(b.y-Y,b.x-X)) return a.x<b.x;
    return (atan2(a.y-Y,a.x-X))<(atan2(b.y-Y,b.x-X));
}
void graham()
{
    int i;
    sort(e+1,e+n+1,cmp);
    top=-1;
    stk[++top]=e[1];
    X=e[1].x;Y=e[1].y;
    sort(e+2,e+n+1,cmp2);
    for (i=2;i<=n;i++)
    {
        while (top>1 && cross(stk[top-1],stk[top],e[i])<eps) top--;
        stk[++top]=e[i];
    }
}
int main()
{
    int i,j;
    double dist,ans=0;
    scanf("%d",&n);
    for (i=1;i<=n;i++) scanf("%lf%lf",&e[i].x,&e[i].y);
    graham();
    stk[++top]=stk[0];
    j=1;
    for (i=0;i<top;i++)
    {
        while (cross(stk[i+1],stk[j],stk[i])<cross(stk[i+1],stk[j+1],stk[i])) j=(j+1)%top;
        dist=getdis(stk[i],stk[j]);
        if (dist>ans) ans=dist;
        dist=getdis(stk[i+1],stk[j+1]);
        if (dist>ans) ans=dist;
    } 
    printf("%.0lf
",ans);
    return 0;
} 

A    LibreOJ 2008

C    CodeForces 1143F

把每个点 的坐标转化为(x,y-x²)，函数转化为y-x²=bx+c。这样就把抛物线转化为了直线，原问题就变成了有多少条直线满足上方没有点。

求n个点中上凸包上的线段数量（点数-1）即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100005
using namespace std;
struct node
{
    double x,y;
}e[maxn],stk[maxn];
bool cmp(node a,node b)
{
    if (a.x==b.x) return a.y<b.y;
    return a.x<b.x;
}
double cross(node a,node b,node c)
{
    node ab,ac;
    ab.x=b.x-a.x;
    ab.y=b.y-a.y;
    ac.x=c.x-a.x;
    ac.y=c.y-a.y;
    return ab.x*ac.y-ab.y*ac.x;
}
double getdis(node a,node b)
{
    node ab;
    ab.x=b.x-a.x;
    ab.y=b.y-a.y;
    return sqrt(ab.x*ab.x+ab.y*ab.y);
}
int main()
{
    int i,n,top;
    double ans;
    scanf("%d",&n);
    for (i=1;i<=n;i++) 
    {
        scanf("%lf%lf",&e[i].x,&e[i].y);
        e[i].y-=e[i].x*e[i].x;
    }
    sort(e+1,e+n+1,cmp);
    top=0;
    for (i=1;i<=n;i++)
    {
        while (top>1 && cross(stk[top-1],stk[top],e[i])>=0) top--;
        stk[++top]=e[i];
    }
    if (stk[1].x==stk[2].x) top--;
    printf("%d
",top-1); 
    return 0;
} 

D    POJ 1259