POJ 1385 Lifting the Stone 计算多边形的重心

本文介绍了一种计算给定多边形重心的方法,通过将多边形分割为三角形并求其加权平均来实现。文章提供了一个具体的C++代码实现,展示了如何处理大量输入点并避免空间限制。

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http://poj.org/problem?id=1385

There are many secret openings in the floor which are covered by a big heavy stone. When the stone is lifted up, a special mechanism detects this and activates poisoned arrows that are shot near the opening. The only possibility is to lift the stone very slowly and carefully. The ACM team must connect a rope to the stone and then lift it using a pulley. Moreover, the stone must be lifted all at once; no side can rise before another. So it is very important to find the centre of gravity and connect the rope exactly to that point. The stone has a polygonal shape and its height is the same throughout the whole polygonal area. Your task is to find the centre of gravity for the given polygon. 

Input
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer N (3 <= N <= 1000000) indicating the number of points that form the polygon. This is followed by N lines, each containing two integers Xi and Yi (|Xi|, |Yi| <= 20000). These numbers are the coordinates of the i-th point. When we connect the points in the given order, we get a polygon. You may assume that the edges never touch each other (except the neighboring ones) and that they never cross. The area of the polygon is never zero, i.e. it cannot collapse into a single line.
Output
Print exactly one line for each test case. The line should contain exactly two numbers separated by one space. These numbers are the coordinates of the centre of gravity. Round the coordinates to the nearest number with exactly two digits after the decimal point (0.005 rounds up to 0.01). Note that the centre of gravity may be outside the polygon, if its shape is not convex. If there is such a case in the input data, print the centre anyway.
Sample Input

2
4
5 0
0 5
-5 0
0 -5
4
1 1
11 1
11 11
1 11

Sample Output

0.00 0.00
6.00 6.00

题目大意:计算给定的多边形的重心。
思路:将多边形分割为三角形的并,并对每个三角形求重心(三角形重心即为三点坐标的平均值),然后以三角形的邮箱面积为权值求加权平均即可。在多边形面积为 0 0 0时重心时没有定义的,但是本题保证了面积不为 0 0 0。但是这道题时卡空间的,因此就不能用数组把点全部存下来了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;

const double pi=acos(-1.0);//弧度pi
const double eps=1e-10;//精度

struct point
{
    double x,y;
    point(double a=0,double b=0)
    {
        x=a,y=b;
    }
    friend point operator * (point a,double b)
    {
        return point(a.x*b,a.y*b);
    }
    friend point operator * (double a,point b)
    {
        return point(b.x*a,b.y*a);
    }
    point operator - (const point &b)const
    {
        return point(x-b.x,y-b.y);
    }
    point operator + (const point &b)const
    {
        return point(x+b.x,y+b.y);
    }
    point operator / (const double b)const
    {
        return point(x/b,y/b);
    }
    bool operator < (const point &b)const//按坐标排序
    {
        if(fabs(x-b.x)<eps)
            return y<b.y-eps;
        return x<b.x-eps;
    }
    void transxy(double sinb,double cosb)//逆时针旋转b弧度
    {                                      //若顺时针 在传入的sinb前加个-即可
        double tx=x,ty=y;
        x=tx*cosb-ty*sinb;
        y=tx*sinb+ty*cosb;
    }
    void transxy(double b)//逆时针旋转b弧度
    {                     //若顺时针传入-b即可
        double tx=x,ty=y;
        x=tx*cos(b)-ty*sin(b);
        y=tx*sin(b)+ty*cos(b);
    }
    double norm()
    {
        return sqrt(x*x+y*y);
    }
};

inline double dot(point a,point b)//点积
{
    return a.x*b.x+a.y*b.y;
}
inline double cross(point a,point b)//叉积
{
    return a.x*b.y-a.y*b.x;
}

inline double dist(point a,point b)//两点间距离
{
    return (a-b).norm();
}

inline int sgn(double x)
{
    if(fabs(x)<eps)
        return 0;
    if(x>0)
        return 1;
    return -1;
}

int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}

bool point_on_seg(point p,point s,point t)//判断点p 是否在线段st上 包括端点
{
    return sgn(cross(p-s,t-s))==0&&sgn(dot(p-s,p-t))<=0;
}

typedef point Vector;

const int maxn=1e6+5;
int n;

int main()
{
    int t;
    scanf("%d",&t);
    point first,pre,cur;
    while(t--)
    {
        scanf("%d",&n);
        scanf("%lf%lf",&first.x,&first.y);
        pre.x=first.x,pre.y=first.y;
        point ans=point(0,0);
        double sum=0;
        for(int i=1;i<n;i++)
        {
            scanf("%lf%lf",&cur.x,&cur.y);
            ans=ans+(pre+cur)*cross(cur,pre);
            sum+=cross(cur,pre);
            pre=cur;
        }
        ans=ans+(first+cur)*cross(first,cur);
        sum+=cross(first,cur);
        sum/=2;
        ans=ans/sum/6;
        if(sgn(ans.x)==0)
            ans.x=0;
        if(sgn(ans.y)==0)
            ans.y=0;
        printf("%.2f %.2f\n",ans.x,ans.y);
    }
    return 0;
}

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