hdu3685（几何重心与凸包结合）

来源：互联网发布：30岁后的美工编辑：程序博客网时间：2024/05/01 00:13

题意：给一个多边形（有可能是凹多边形）。问有多少种能够使得它稳定放置的方式。当然稳定的原则就是重心做垂线在支撑点之内。

解法：因为有可能是凹多边形，所以先求出多边形的凸包，这是在放置时候会接触地面的所有点。然后将重心与每天凸边判断是否稳定；

代码：

/******************************************************* @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;}

0 0