基於水平序的Andrew凸包演算法 最詳細的圖解(多圖預警)
阿新 • • 發佈:2019-01-03
給出凸包的定義:
簡要說一下思路:
首先將所有點按照x從小到大(x同則y從小到大)排序
把p1,p2放入凸包,從p3開始,當新點在凸包‘前進’方向的左邊時繼續,否則依次刪除最近加入凸包的點,直到新點在左邊
輸入不能有重複點,不希望凸包邊上有點可疑將<=改為<
下面請根據程式,測試資料與輸出 及步驟圖學習,跟著模擬一遍就會非常明白了
#include<bits/stdc++.h> using namespace std; const double eps=1e-10; struct Point{ double x,y; int id; Point(double x=0,double y=0,int id=0):x(x),y(y),id(id){} }; typedef Point Vector; bool operator< (const Point&a,Point &b){ return a.x<b.x ||(a.x==b.x && a.y<b.y); } Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);} Vector operator - (Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);} Vector operator * (Vector A,double B){return Vector(A.x*B,A.y*B);} Vector operator / (Vector A,double B){return Vector(A.x/B,A.y/B);} int dcmp(double x){if(fabs(x)<eps)return 0;return (x>0)?1:-1;} bool operator == (const Vector A,const Vector B){ return dcmp(A.x-B.x)==0 && dcmp(A.y-B.y)==0; } double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;} double Length(Vector A){return sqrt(Dot(A,A));} double Cross(Vector A,Vector B){return A.x*B.y-B.x*A.y;} int ConvexHell(Point p[],int n,Point 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){ //下凸包 printf("update: segment %d TO %d destination: %d\n",m-2,m-1,i); m--; } ch[m++]=p[i]; printf("ch[ %d ]= %d\n",m-1,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){ //上凸包 printf("update: segment %d TO %d destination: %d\n",m-2,m-1,i); m--; } ch[m++]=p[i]; printf("ch[ %d ]= %d\n",m-1,i); } if(n>1)m--; return m; } int main(){ int n; scanf("%d",&n); Point p[20],ch[20]; for(int i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y); int sum=ConvexHell(p,n,ch); printf("Final Ans:\n"); for(int i=0;i<sum;i++)printf("%.0lf %.0lf,",ch[i].x,ch[i].y); return 0; }
/*測試資料及輸出 8 3 0 4 8 5 5 6 3 6 7 8 4 9 2 10 7 ch[ 0 ]= 0 ch[ 1 ]= 1 update: segment 0 TO 1 destination: 2 ch[ 1 ]= 2 update: segment 0 TO 1 destination: 3 ch[ 1 ]= 3 ch[ 2 ]= 4 update: segment 1 TO 2 destination: 5 update: segment 0 TO 1 destination: 5 ch[ 1 ]= 5 update: segment 0 TO 1 destination: 6 ch[ 1 ]= 6 ch[ 2 ]= 7 ch[ 3 ]= 6 update: segment 2 TO 3 destination: 5 ch[ 3 ]= 5 update: segment 2 TO 3 destination: 4 ch[ 3 ]= 4 ch[ 4 ]= 3 update: segment 3 TO 4 destination: 2 ch[ 4 ]= 2 update: segment 3 TO 4 destination: 1 update: segment 2 TO 3 destination: 1 ch[ 3 ]= 1 ch[ 4 ]= 0 Final Ans: 3 0,9 2,10 7,4 8, */