uva10256(计算几何)

省选前练模板系列;

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=533;
const double eps=1e-10,inf=1e9;
int n1,n2,m1,m2,pp;
double mi;
int dcmp(double x){if(fabs(x)<eps)return 0;return x<0?-1:1;}
struct vec{
    double x,y,ang;
    vec(){};
    vec(double a,double b){x=a,y=b;}
    vec operator-(vec a){return vec(x-a.x,y-a.y);}
    vec operator+(vec a){return vec(x+a.x,y+a.y);}
    bool operator<(const vec&a)const{
        return dcmp(x-a.x)<0||(dcmp(x-a.x)==0&&y<a.y);
    }
}p1[maxn],p2[maxn],ch1[maxn],ch2[maxn],p0,t1,t2,t3;
double cross(vec a,vec b){return a.x*b.y-b.x*a.y;}
double dot(vec a,vec b){return a.x*b.x+a.y*b.y;}
bool onli(vec p,vec a1,vec a2){
    if(p.x=a1.x&&p.y==a1.y)return 1;
    if(p.x=a2.x&&p.y==a2.y)return 1;
    return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;
}
bool jiao(vec a1,vec a2,vec b1,vec b2){
    if(!(min(a1.x,a2.x)<=max(b1.x,b2.x)&&min(b1.y,b2.y)<=max(a1.y,a2.y)&&min(b1.x,b2.x)<=max(a1.x,a2.x)&&min(a1.y,a2.y)<=max(b1.y,b2.y)))
    return 0;
    double c1=cross(a2-a1,b1-a1),c2=cross(a2-a1,b2-a1);
    double c3=cross(b2-b1,a1-b1),c4=cross(b2-b1,a2-b1);
    return dcmp(c1)*dcmp(c2)<=0&&dcmp(c3)*dcmp(c4)<=0;
}
bool inbao(vec p,vec* a,int n){
    if(dcmp(cross(a[1]-a[0],p-a[0]))<0||dcmp(cross(a[n-1]-a[0],p-a[0]))>0)return 0;
    if(dcmp(cross(a[1]-a[0],p-a[0]))==0){
        if(onli(p,a[0],a[1]))return 1;
        else return 0;
    }
    if(dcmp(cross(a[n-1]-a[0],p-a[0]))==0){
        if(onli(p,a[0],a[n-1]))return 1;
        else return 0;
    }
    int l=2,r=n-1;
    while(l<r){
        int mid=l+r>>1;
        if(dcmp(cross(p-a[0],a[mid]-a[0]))>0)r=mid;
        else l=mid+1;
    }
    //cout<<l<<endl;
    if(cross(a[l]-a[l-1],p-a[l-1])>=0)return 1;
    return 0;
}
int build(vec *p,int n,vec *ch){
    sort(p,p+n);
    int m=0;
    for(int i=0;i<n;++i){
        while(m>1&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
        ch[m++]=p[i];
    }
    int k=m;
    for(int i=n-2;i>=0;--i){
        while(m>k&&cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
        ch[m++]=p[i];
    }
    if(n>1)m--;
    return m;
}
void solve(){
    for(int i=0;i<n1;++i)scanf("%lf%lf",&p1[i].x,&p1[i].y);
    for(int i=0;i<n2;++i)scanf("%lf%lf",&p2[i].x,&p2[i].y);
    if(n1==1&&n2==1){
        if(p1[0].x==p2[0].x&&p1[0].y==p2[0].y){puts("No");return;}
        else{puts("Yes");return;}
    }
    m1=build(p1,n1,ch1),m2=build(p2,n2,ch2);
    if(m2>2){
        for(int i=0;i<m1;++i)if(inbao(ch1[i],ch2,m2)){puts("No");return;}
    }
    if(m1>2){
        for(int i=0;i<m2;++i)if(inbao(ch2[i],ch1,m1)){puts("No");return;}
    }
    for(int i=0;i<m1;++i)
        for(int j=0;j<m2;++j)
        if(jiao(ch1[i],ch1[(i+1)%m1],ch2[j],ch2[(j+1)%m2])){
              puts("No");return;
    }
    puts("Yes");
}
int main(){
    while(scanf("%d%d",&n1,&n2)!=EOF&&n1){
        solve();
    }
    return 0;
}
/*
4 3 
100 600 
200 400 
600 500 
300 700 
400 100 
600 200 
500 300 
4 3 
100 600 
400 100 
600 200 
500 300 
200 400 
600 500 
300 700 
0 0
*/
原文地址:https://www.cnblogs.com/dibaotianxing/p/8711677.html