HDU 1115 Lifting the Stone （求多边形的重心：叉积）

最新推荐文章于 2019-12-04 12:26:39 发布

原创最新推荐文章于 2019-12-04 12:26:39 发布 · 542 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#HDU #1115 #Lifting the Stone #求多边形的重心 #叉积

HDUOJ 同时被 2 个专栏收录

304 篇文章

订阅专栏

ACM_多边形

9 篇文章

订阅专栏

本文介绍了一种计算多边形重心的方法，通过将多边形分解为多个三角形，求得各三角形的重心坐标及面积，进而计算出整个多边形的重心坐标。

Lifting the Stone

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7328 Accepted Submission(s): 3088

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

Sample Output

  
   0.00 0.00
6.00 6.00

Source

Central Europe 1999

Recommend

Eddy | We have carefully selected several similar problems for you: 1392 1086 2108 2150 1348

题意：求多边形的重心。

题解：

1、把多边形划分成多个三角形，求每个三角形的重心，重心坐标为：（(x0 + x1 + x2) / 3，（y0 + y1 + y2) / 3)，三角形面积用叉积求，对凸多边形和凹多边形都适用（因为值有正负）；

2、作为二维的多边形，把面积作为权值，分别乘以重心坐标的X和Y值；

3、分别将求出的X, Y值的加权平均数除以总面积，即多边形面积的重心坐标。

AC代码：

#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int cas , n ;
struct center
{
       double x , y ;
};
double Area( center p0 , center p1 , center p2 )//叉积求面积
{
       double area = 0 ;
       area =  p0.x * p1.y + p1.x * p2.y + p2.x * p0.y - p1.x * p0.y - p2.x * p1.y - p0.x * p2.y;
       return area / 2 ;  
}
int main ()
{
    center p0 , p1 , p2 ;
    double sum_x , sum_y , sum_area , area;
    scanf ( "%d" , &cas ) ;
    while ( cas -- )
    {
          sum_x = sum_y = sum_area = 0 ;
          scanf ( "%d" , &n ) ;
          scanf ( "%lf%lf" , &p0.x , &p0.y ) ;
          scanf ( "%lf%lf" , &p1.x , &p1.y ) ;
          
          for ( int i = 2 ; i < n ; ++ i )
          {
              scanf ( "%lf%lf" , &p2.x , &p2.y ) ;
              area = Area(p0,p1,p2) ;
              sum_area += area ;
              sum_x += (p0.x + p1.x + p2.x) * area ;
              sum_y += (p0.y + p1.y + p2.y) * area ;
              p1 = p2 ;
          }
          printf ( "%.2lf %.2lf\n" , sum_x / sum_area / 3 , sum_y / sum_area / 3 ) ;
    }
    return 0 ;
}