Hdu3685 (combination of geometric center of gravity and convex hull) and hdu3685 convex hull

Hdu3685 (combination of geometric center of gravity and convex hull) and hdu3685 convex hull

Question: Give a polygon (it may be a concave polygon ). Ask how many methods can be used to make it stable. Of course, the principle of stability is that the center of gravity is laid down within the Support Point.

Solution: because it may be a concave polygon, first obtain the convex hull of the polygon. This is all points that will touch the ground when placed. Then the center of gravity and the convex side of the day are determined to be stable;

Code:

/******************************************************* @author:xiefubao*******************************************************/#pragma comment(linker, "/STACK:102400000,102400000")#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <queue>#include <vector>#include <algorithm>#include <cmath>#include <map>#include <set>#include <string.h>//freopen ("in.txt" , "r" , stdin);using namespace std;#define eps 1e-8#define zero(_) (_<=eps)const double pi=acos(-1.0);typedef long long LL;const int Max=100010;const LL INF=0x3FFFFFFF;struct point{    double x,y;};point points[50005];point focus;int top;int stack[50005];int N;double mult(point a,point b,point c){    a.x-=c.x;    a.y-=c.y;    b.x-=c.x;    b.y-=c.y;    return a.x*b.y-a.y*b.x;}double dis(const point& a,const point& b){    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}bool operator<(point a,point b){    if(mult(a,b,points[0])>0||(mult(a,b,points[0])==0 && a.x<b.x))        return true;    else        return false;}point OK(point a,point b,point c){    point ans;    ans.x=(a.x+b.x+c.x)/3;    ans.y=(a.y+b.y+c.y)/3;    return ans;}void getFocus(point& focus,point* points,int N){    focus.x=0;    focus.y=0;    double base=0;    for(int i=2; i<N; i++)    {        double t=mult(points[0],points[i-1],points[i]);        point pp=OK(points[0],points[i-1],points[i]);        focus.x+=t*pp.x;        focus.y+=t*pp.y;        base+=t;    }    focus.x/=base;    focus.y/=base;}int getans(){    int ans=0;    for(int i=0; i<top; i++)    {        double l=dis(focus,points[stack[i]]);        double r=dis(focus,points[stack[i+1]]);        double u=dis(points[stack[i]],points[stack[i+1]]);        if(l+u>r&&r+u>l)            ans++;    }    double l=dis(focus,points[stack[top]]);    double r=dis(focus,points[stack[0]]);    double u=dis(points[stack[top]],points[stack[0]]);    if(l+u>r&&r+u>l)        ans++;    return ans;}void graham(int n){    int mi=0;    for(int i=1; i<n; i++)    {        if(points[i].y<points[mi].y||(points[i].y==points[mi].y&&points[i].x<points[mi].x))            mi=i;    }    point a=points[0];    points[0]=points[mi];    points[mi]=a;    sort(points+1,points+n);    stack[0]=0;    stack[1]=1;    stack[2]=2;    top=2;    for(int i=3; i<n; i++)    {        while(top>0&&mult(points[stack[top]],points[stack[top-1]],points[i])>=0)        {            top--;        }        stack[++top]=i;    }}int main(){    int t;    cin>>t;    while(t--)    {        scanf("%d",&N);        for(int i=0; i<N; i++)        {            scanf("%lf%lf",&points[i].x,&points[i].y);        }        getFocus();        graham(N);        cout<<getans()<<endl;    }    return 0;}

What is the geometric center and center of gravity? Different? Contact ?? Dizzy ,?

Only a regular image has a geometric center, such as a square or a positive triangle. Each ry has a ry center of gravity.

The ry center of gravity is located at the ry center when it is a regular ry of uniform media.

We have just completed the course today. =

The difference between the ry center and the ry Center

Only a regular image has a geometric center, such as a square or a positive triangle. Each ry has a ry center of gravity (for example, a triangle is the intersection of three midline). When it is a regular ry, The ry center is in the ry center.