多邊形重心問(wèn)題

330 10 20 331 10 00 141 10 00 0.50 1樣例輸出
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質(zhì)量集中在頂點(diǎn)上* n個(gè)頂點(diǎn)坐標(biāo)為(xi,yi)，質(zhì)量為mi，則重心*　X = ∑( xi×mi ) / ∑mi*　Y = ∑( yi×mi ) / ∑mi*　特殊地，若每個(gè)點(diǎn)的質(zhì)量相同，則*　X = ∑xi / n*　Y = ∑yi / n*②質(zhì)量分布均勻*　特殊地，質(zhì)量均勻的三角形重心：*　X = ( x0 + x1 + x2 ) / 3*　Y = ( y0 + y1 + y2 ) / 3*③三角形面積公式：S = ( (x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1) ) / 2 ;*做題步驟：1、將多邊形分割成n-2個(gè)三角形，根據(jù)③公式求每個(gè)三角形面積。* 2、根據(jù)②求每個(gè)三角形重心。* 3、根據(jù)①求得多邊形重心。**/
兩種方法：
一種是將n個(gè)點(diǎn)，以其中一個(gè)點(diǎn)為標(biāo)準(zhǔn)，分成n-2個(gè)三角形，再進(jìn)行求重心。
另一種是以原點(diǎn)為依據(jù)分成n+1個(gè)三角形，再進(jìn)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 Input
245 00 5-5 00 -541 111 111 111 11 Sample Output
0.00 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;}