多邊形重心問題
時間限制：3000 ms | 內存限制：65535 KB難度：5描述在某個多邊形上，取n個點，這n個點順序給出，按照給出順序將相鄰的點用直線連接， （第一個和最后一個連接），所有線段不和其他線段相交，但是可以重合，可得到一個多邊形或一條線段或一個多邊形和一個線段的連接后的圖形； 如果是一條線段,我們定義面積為0，重心坐標為（0,0）.現在求給出的點集組成的圖形的面積和重心橫縱坐標的和；輸入第一行有一個整數0<n<11,表示有n組數據；每組數據第一行有一個整數m<10000,表示有這個多邊形有m個頂點；輸出輸出每個多邊形的面積、重心橫縱坐標的和，小數點后保留三位；樣例輸入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質量集中在頂點上* n個頂點坐標為(xi,yi)，質量為mi，則重心* X = ∑( xi×mi ) / ∑mi* Y = ∑( yi×mi ) / ∑mi* 特殊地，若每個點的質量相同，則* X = ∑xi / n* Y = ∑yi / n*②質量分布均勻* 特殊地，質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個三角形，根據③公式求每個三角形面積。* 2、根據②求每個三角形重心。* 3、根據①求得多邊形重心。**/
兩種方法：
一種是將n個點，以其中一個點為標準，分成n-2個三角形，再進行求重心。
另一種是以原點為依據分成n+1個三角形，再進行求重心。
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;}
