多边形重心问题
时间限制：3000 ms | 内存限制：65535 KB难度：5描述在某个多边形上，取n个点，这n个点顺序给出，按照给出顺序将相邻的点用直线连接， （第一个和最后一个连接），所有线段不和其他线段相交，但是可以重合，可得到一个多边形或一条线段或一个多边形和一个线段的连接后的图形； 如果是一条线段,我们定义面积为0，重心坐标为（0,0）.现在求给出的点集组成的图形的面积和重心横纵坐标的和；输入第一行有一个整数0<n<11,表示有n组数据；每组数据第一行有一个整数m<10000,表示有这个多边形有m个顶点；输出输出每个多边形的面积、重心横纵坐标的和，小数点后保留三位；样例输入330 10 20 331 10 00 141 10 00 0.50 1样例输出0.000 0.0000.500 1.0000.500 1.000#include<cstdio>int cases,n;double a[10002],b[10002],t,xx,yy,area;inline double fun(int i){ return (a[i]*b[i+1]-a[i+1]*b[i])/2;}inline double ffabs(double i){ return i>0?i:-i;}int main(){ scanf("%d",&cases); while(cases--){ xx=yy=area=0; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%lf%lf",&a[i],&b[i]); a[n+1]=a[1],b[n+1]=b[1]; for(int i=1;i<=n;i++){ t=fun(i); area+=t; xx+=t*(a[i]+a[i+1]); yy+=t*(b[i]+b[i+1]); } xx=xx/3/area; yy=yy/3/area; area=ffabs(area); if(area<1e-8) PRintf("0.000 0.000/n"); else printf("%.3lf %.3lf/n",area,xx+yy);}return 0;}*①质量集中在顶点上* n个顶点坐标为(xi,yi)，质量为mi，则重心* X = ∑( xi×mi ) / ∑mi* Y = ∑( yi×mi ) / ∑mi* 特殊地，若每个点的质量相同，则* X = ∑xi / n* Y = ∑yi / n*②质量分布均匀* 特殊地，质量均匀的三角形重心：* X = ( x0 + x1 + x2 ) / 3* Y = ( y0 + y1 + y2 ) / 3*③三角形面积公式：S = ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2 ;*做题步骤：1、将多边形分割成n-2个三角形，根据③公式求每个三角形面积。* 2、根据②求每个三角形重心。* 3、根据①求得多边形重心。**/
两种方法：
一种是将n个点，以其中一个点为标准，分成n-2个三角形，再进行求重心。
另一种是以原点为依据分成n+1个三角形，再进行求重心。
Lifting the Stone
Time Limit: 2000/1000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7885 Accepted Submission(s): 3331Problem DescriptionThere 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. InputThe 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. OutputPrint 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 Input245 00 5-5 00 -541 111 111 111 11 Sample Output0.00 0.006.00 6.00 SourceCentral Europe 1999#include<cstdio>int cases,n;double a[10002],b[10002],t,xx,yy,area;inline double fun(int i){ return (a[i]*b[i+1]-a[i+1]*b[i])/2;}inline double ffabs(double i){ return i>0?i:-i;}int main(){ scanf("%d",&cases); while(cases--){ xx=yy=area=0; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%lf%lf",&a[i],&b[i]); a[n+1]=a[1],b[n+1]=b[1]; for(int i=1;i<=n;i++){ t=fun(i); area+=t; xx+=t*(a[i]+a[i+1]); yy+=t*(b[i]+b[i+1]); } xx=xx/3/area; yy=yy/3/area; printf("%.2lf %.2lf/n",xx,yy);}return 0;}
