几何求叉积+最短路——cf1032D

直接无脑求交点然后求最短路了

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

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};}
    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};}
}; 
db cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
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 checkSL(point k1,point k2,point k3,point k4){//线段和直线严格相交
    int sign1 = sign(cross(k3-k1,k4-k1));
    int sign2 = sign(cross(k3-k2,k4-k2));
    return sign(cross(k3-k1,k4-k1))*sign(cross(k3-k2,k4-k2))<0; 
}
point getSL(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);
}
point p[10],k1,k2;
db a,b,c,x1,y1,x2,y2,dis[10][10],d[10];
int tot,v[10];
void dij(){
    for(int i=1;i<=6;i++)d[i]=1e16,v[i]=0;
    d[1]=0;
    for(int i=1;i<=6;i++){
        int now;db Min=1e16;
        for(int j=1;j<=6;j++)
            if(Min>d[j] && v[j]==0){
                Min=d[i];now=j;
            }
        v[now]=1;
        for(int j=1;j<=6;j++)
            if(!v[j])d[j]=min(d[j],d[now]+dis[now][j]); 
    }
}
//ax+by+c=0
int main(){
    cin>>a>>b>>c>>x1>>y1>>x2>>y2;
    p[1]=(point){x1,y1};p[2]=(point){x1,y2};
    p[3]=(point){x2,y2};p[4]=(point){x2,y1};
    if(a==0){
        k1=(point){0,-c/b};
        k2=(point){1,-c/b};
    }else if(b==0){
        k1=(point){-c/a,0};
        k2=(point){-c/a,1};
    }else {
        k1=(point){0,-c/b};
        k2=(point){1,(-c-a)/b};
    }
    
    for(int i=1;i<=6;i++)
        for(int j=1;j<=6;j++)
            dis[i][j]=1e16;
            
    tot=4;
    if(sign(a*p[1].x+b*p[1].y+c)==0){
        p[++tot]=p[1];
        dis[1][tot]=dis[tot][1]=0;
    }
    if(sign(a*p[2].x+b*p[2].y+c)==0){
        p[++tot]=p[2];
        dis[2][tot]=dis[tot][2]=0;
    }
    if(sign(a*p[3].x+b*p[3].y+c)==0){
        p[++tot]=p[3];
        dis[3][tot]=dis[tot][3]=0;
    }
    if(sign(a*p[4].x+b*p[4].y+c)==0){
        p[++tot]=p[4];
        dis[4][tot]=dis[tot][4]=0;
    }
    if(checkSL(p[1],p[2],k1,k2)){
        point k3=getSL(p[1],p[2],k1,k2);
        p[++tot]=k3;
        dis[1][tot]=dis[tot][1]=k3.dis(p[1]);
        dis[2][tot]=dis[tot][2]=k3.dis(p[2]);
    }
    if(checkSL(p[2],p[3],k1,k2)){
        point k3=getSL(p[2],p[3],k1,k2);
        p[++tot]=k3;
        dis[3][tot]=dis[tot][3]=k3.dis(p[3]);
        dis[2][tot]=dis[tot][2]=k3.dis(p[2]);
    }
    if(checkSL(p[3],p[4],k1,k2)){
        point k3=getSL(p[3],p[4],k1,k2);
        p[++tot]=k3;
        dis[3][tot]=dis[tot][3]=k3.dis(p[3]);
        dis[4][tot]=dis[tot][4]=k3.dis(p[4]);
    }
    if(checkSL(p[4],p[1],k1,k2)){
        point k3=getSL(p[4],p[1],k1,k2);
        p[++tot]=k3;
        dis[1][tot]=dis[tot][1]=k3.dis(p[1]);
        dis[4][tot]=dis[tot][4]=k3.dis(p[4]);
    }
    dis[1][2]=dis[2][1]=p[1].dis(p[2]);
    dis[2][3]=dis[3][2]=p[2].dis(p[3]);
    dis[3][4]=dis[4][3]=p[3].dis(p[4]);
    dis[4][1]=dis[1][4]=p[1].dis(p[4]);
    dis[5][6]=dis[6][5]=p[5].dis(p[6]);
    
    if(tot==4){
        printf("%.10lf
",fabs(x1-x2)+fabs(y1-y2));
    }
    else {
        dij();
        printf("%.10lf
",d[3]);
    }
}