【经典】凸包+hash+kmp——cf1017E

 求两个凸包，问其中一个能否在平移，旋转之后与第二个重合

把凸包用 边-角-边 这样的形式hash出来，然后用字符串匹配的方式进行匹配即可

#include<bits/stdc++.h>
using namespace std;
#define N 200005
#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;}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
    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;
}
vector<point> ConvexHull(vector<point>A,int flag=1){
    int n=A.size(); vector<point>ans(n*2); 
    sort(A.begin(),A.end()); int now=-1;
    for (int i=0;i<A.size();i++){
        while (now>0&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
        ans[++now]=A[i];
    } int pre=now;
    for (int i=n-2;i>=0;i--){
        while (now>pre&&sign(cross(ans[now]-ans[now-1],A[i]-ans[now-1]))<flag) now--;
        ans[++now]=A[i];
    } ans.resize(now); return ans;
}

vector<point> v1,v2,con1,con2;
int n,m;

struct Node{
    db len1,len2,cross;
    bool operator!=(const Node & a)const {
        if(sign(len1-a.len1)!=0 || sign(len2-a.len2)!=0 || sign(cross-a.cross)!=0)return 1;
        return 0;
    }
}s1[N],s2[N];

int nxt[N];
void kmp_pre(){
    int i,j;
    int len=con2.size();
    j=nxt[0]=-1;i=0;
    while(i<len){
        while(j!=-1 && s2[i]!=s2[j])j=nxt[j];
            nxt[++i]=++j;
    }
}
void kmp(){
    int i,j;
    int m=con2.size();
    int n=con1.size()*2;
    i=j=0;
    while(i<n){
        while(j!=-1 && s1[i]!=s2[j])j=nxt[j];
        ++i,++j;
        if(j==m){
            puts("YES");return;
        }
    }
    puts("NO");
}

int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        point k;
        scanf("%lf%lf",&k.x,&k.y);
        v1.push_back(k);
    }
    for(int i=1;i<=m;i++){
        point k;scanf("%lf%lf",&k.x,&k.y);
        v2.push_back(k);
    }
    
    con1=ConvexHull(v1,1);
    con2=ConvexHull(v2,1);
    
    if(con1.size()!=con2.size()){puts("NO");return 0;}
    
    for(int i=0;i<con1.size();i++){
        Node cur;
        point k1=con1[i],k2=con1[(i+1)%con1.size()],k3=con1[(i+2)%con1.size()];
        cur.len1=k1.dis(k2);
        cur.len2=k2.dis(k3);
        cur.cross=dot(k1-k2,k3-k2);
        s1[i]=cur;
    }
    for(int i=0;i<con2.size();i++){
        Node cur;
        point k1=con2[i],k2=con2[(i+1)%con1.size()],k3=con2[(i+2)%con1.size()];
        cur.len1=k1.dis(k2);
        cur.len2=k2.dis(k3);
        cur.cross=dot(k1-k2,k3-k2);
        s2[i]=cur;
    }
    
    /*for(int i=0;i<con1.size();i++)
        cout<<con1[i].x<<" "<<con1[i].y<<'
';
    //    cout<<s1[i].len1<<" "<<s1[i].len2<<" "<<s1[i].cross<<"
"; 
    for(int i=0;i<con1.size();i++)
        cout<<con2[i].x<<" "<<con2[i].y<<'
';
    //    cout<<s2[i].len1<<" "<<s2[i].len2<<" "<<s2[i].cross<<"
"; 
    */
    kmp_pre(); 
    for(int i=0;i<con1.size();i++)
        s1[i+con1.size()]=s1[i];
    kmp();
}