2/14 preliminary calculation geometry

Properties of convex polygons ： It means that every interior angle of a polygon is less than 180180 Degree polygon
Turning the vector to the right means that the image is recessed
https://www.luogu.com.cn/problem/P3744
Make a point in a convex polygon move the least distance to make it become a concave polygon

#include <bits/stdc++.h>
#define inf 1000000000.0
using namespace std;

struct node
{
    
    double x,y;
    node (double x=0,double y=0):x(x),y(y){
    }
    node operator- (const node &r) const
    {
    
        return node{
    x-r.x,y-r.y};
    }
}e[1005];
int n;
double ans=inf;
double cross(node e1,node e2)
{
    
    return fabs(e1.x*e2.y-e1.y*e2.x);
}
double len(node e1,node e2)
{
    
    return sqrt((e1.x-e2.x)*(e1.x-e2.x)+(e1.y-e2.y)*(e1.y-e2.y));
}
double go(int i)
{
    
    node e1=e[i]-e[i+1],e2=e[i]-e[i+2];
    return cross(e1,e2)/len(e2,node(0.0));
}
signed main()
{
    
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
    
        scanf("%lf%lf",&e[i].x,&e[i].y);
    }
    e[n+1]=e[1];e[n+2]=e[2];
    for(int i=1;i<=n;i++)
        ans=min(ans,go(i));
    cout<<ans/2<<endl;
    return 0;
}

Simple application of cross product
https://www.luogu.com.cn/problem/P1355
Notice the order of the coordinates and the direction of the vector

#include <bits/stdc++.h>

using namespace std;

int read()
{
    
    int ret = 0, ch;
    while(!isdigit(ch = getchar()) && ch != '-');
    bool bm = (ch == '-'); if(bm) ch = getchar();
    while(ret = ret * 10 + (ch - '0'), isdigit(ch = getchar()));
    return bm ? -ret : ret;
}
struct node
{
    
    int x,y;
    bool operator==(const node &r) const
    {
    
        return x == r.x && y == r.y;
    }
    node operator-(const node &r) const
    {
    
        return (node){
    x-r.x,y-r.y};
    }
}a,b,c,p,q;

int cross(const node &a,const node &b)  // Cross product 
{
    
    return a.x*b.y-a.y*b.x;
}

bool inn(const node &a,const node &b,const node &c)
{
       // Judge on the side 
    return a.x>=min(b.x,c.x)&&a.x<=max(b.x,c.x)
        && a.y>=min(b.y,c.y)&&a.y<=max(b.y,c.y)
        && cross(a-b,a-c)==0;
}
bool innn(const node &a,const node &b,const node &c,const node &d)
{
       // Judge whether it is inside 
    return (cross(d-c,a-d)^cross(d-c,b-d))>=0;
    // If the vector AB× vector BP And vector AB× vector BH Same number , be q,P stay AB Ipsilateral .
}
signed main()
{
    
    a.x=read()*3;a.y=read()*3;
    b.x=read()*3;b.y=read()*3;
    c.x=read()*3;c.y=read()*3;
    p.x=read()*3;p.y=read()*3;
    q.x=(a.x+b.x+c.x)/3;
    q.y=(a.y+b.y+c.y)/3;
    
    if(p==a||p==b||p==c)
        cout<<"4"<<endl;
        
    else if(inn(p,a,b)||inn(p,b,c)||inn(p,c,a))
        cout<<"3"<<endl;
        
    else if(innn(p,q,a,b)&&innn(p,q,b,c)&&innn(p,q,c,a))
        cout<<"1"<<endl;
    else
        cout<<"2"<<endl;
    return 0;
}

vector a=(x1,y1) and b=(x2,y2) Collinear , available x1y2-x2y1==0
If two points are on same straight line , be x1y2=x2y1

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int maxn=705;
const int inf=0x3f3f3f3f;
struct node
{
    
    int x,y;
}e[maxn];
int n,ma;

signed main()
{
    
    scanf("%lld",&n);
    for(int i=1;i<=n;i++)
    {
    
        scanf("%lld%lld",&e[i].x,&e[i].y);
    }
    for(int i=1;i<=n-1;i++)
    {
    
        for(int j=i+1;j<=n;j++)
        {
    
            int tmp=2;
            node p;
            p.x=e[j].x-e[i].x;p.y=e[j].y-e[i].y;
            for(int g=1;g<=n;g++)
            {
    
                if(g==i||g==j)
                    continue;
                node p1;
                p1.x=e[g].x-e[i].x;p1.y=e[g].y-e[i].y;
                if(p.x*p1.y==p.y*p1.x)
                    tmp++;
            }
            ma=max(ma,tmp);
        }
    }
    cout<<ma<<endl;
    return 0;
}