HDOJ 1115 Lifting the Stone 多邊形重心
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。這個題就是這一類型，算法和上面的不同。
特殊地，質量均勻的三角形重心：
X = ( x0 + x1 + x2 ) / 3
Y = ( y0 + y1 + y2 ) / 3
若我們求出了每個三角形的重心和質量，可以構造一個新的多邊形，頂點為所有三角形的重心，頂點質量為三角形的質量。這個新多邊形的質量和重心與原多邊形相同，即可使用第一種類型的公式計算出整個多邊形的重心。
由於三角形的面積與質量成正比，所以我們這裡用面積代替質量來計算。
現在有個問題就是，多邊形有可能為凹多邊形，三角形有可能在多邊形之外。如何處理這種情況呢？
很簡單，我們使用叉積來計算三角形面積，當三角形在多邊形之外時，得到“負面積”就抵消掉了。
S =( x0*y1 + x1*y2 + x2*y0
- x1*y0 - x2*y1 - x0*y2 ) /2;
Lifting the Stone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5733 Accepted Submission(s): 2400
Problem Description 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
Source Central Europe 1999
#include
#include
#include
#include
using namespace std;
const int maxn=1001000;
struct Point
{
double x,y;
}p[maxn];
int n;
int main()
{
int T_T;
scanf(%d,&T_T);
while(T_T--)
{
scanf(%d,&n);
for(int i=0;i
