计算几何线段判交点+容斥——cf1036E

不知道为什么tag里会有个fft。。

#include<bits/stdc++.h>
using namespace std;
#define N 2005
#define ll long long 
typedef double db;
const db eps=1e-8;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
int inmid(db k1,db k2,db k3){return sign(k1-k3)*sign(k2-k3)<=0;}// k3 在 [k1,k2] 内 
struct point{
    db x,y;
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    int operator == (const point &k1) const{return cmp(x,k1.x)==0&&cmp(y,k1.y)==0;}
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    point unit(){db w=abs(); return (point){x/w,y/w};}
};
int inmid(point k1,point k2,point k3){return inmid(k1.x,k2.x,k3.x)&&inmid(k1.y,k2.y,k3.y);}
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
db dot(point k1,point k2){return k1.x*k2.x+k1.y*k2.y;}
db rad(point k1,point k2){return atan2(cross(k1,k2),dot(k1,k2));} 
point getLL(point k1,point k2,point k3,point k4){
    db w1=cross(k1-k3,k4-k3),w2=cross(k4-k3,k2-k3); return (k1*w2+k2*w1)/(w1+w2);
}
int intersect(db l1,db r1,db l2,db r2){
    if (l1>r1) swap(l1,r1); if (l2>r2) swap(l2,r2); return cmp(r1,l2)!=-1&&cmp(r2,l1)!=-1;
}
int checkSS(point k1,point k2,point k3,point k4){
    return intersect(k1.x,k2.x,k3.x,k4.x)&&intersect(k1.y,k2.y,k3.y,k4.y)&&
    sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<=0&&
    sign(cross(k1-k3,k2-k3))*sign(cross(k1-k4,k2-k4))<=0;
}

point A[N],B[N]; 
ll n,ans;
map<pair<int,int>,int> mp;

int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        int x1,y1,x2,y2;
        cin>>x1>>y1>>x2>>y2;
        A[i]=(point){1.0*x1,1.0*y1};
        B[i]=(point){1.0*x2,1.0*y2};
        ans+=abs(__gcd(x1-x2,y1-y2))+1;
    }
    for(int i=1;i<=n;i++)
        for(int j=i+1;j<=n;j++){
            if(checkSS(A[i],B[i],A[j],B[j])){
                point k=getLL(A[i],B[i],A[j],B[j]);
                pair<int,int> p=make_pair(round(k.x),round(k.y));
                if(sign(p.first-k.x)==0 && sign(p.second-k.y)==0) //判断是整数点 
                    mp[p]++;
            }
        }
    //x条直线经过该点，那么这个点被统计x(x-1)/2次，解出这个x，答案里减去x-1
    for(auto p:mp){
        int cnt=p.second;
        ans-=(1+sqrt(1+8*cnt))/2-1;    
    } 
    cout<<ans<<'\n';
}