Lifting the Stone
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 2121 Accepted Submission(s):
882
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
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Eddy
详情参考 ：《计算几何-多边形重心》
View Code
1 #include<stdio.h>
2 #include<math.h>
3 #include<stdlib.h>
4 #define MAXN 1000005
5 struct centre
6 {
7 double x , y ;
8 }point[MAXN];
9 int cas , n ;
10 struct centre get_centre()
11 {
12 struct centre ctr ;
13 double area = 0 ;
14 ctr.x = ctr.y = 0 ;
15 point[n] = point[0] ;
16 for ( int i = 0 ; i < n ; ++ i )
17 {
18 area +=( ( point[i].x * point[i+1].y - point[i].y * point[i+1].x ) / 2.0 ) ; // 以原点为定点，求三角形面积
19 ctr.x += ( point[i].x * point[i+1].y - point[i].y * point[i+1].x ) * ( point[i].x + point[i+1].x ) ;
20 ctr.y += ( point[i].x * point[i+1].y - point[i].y * point[i+1].x ) * ( point[i].y + point[i+1].y ) ;
21 }
22 ctr.x /= ( area * 6 ) ;
23 ctr.y /= ( area * 6 ) ;
24 return ctr ;
25 }
26 int main ()
27 {
28 scanf ( "%d" , &cas ) ;
29 while ( cas -- )
30 {
31 scanf ( "%d" , &n ) ;
32 for ( int i = 0 ; i < n ; ++ i )
33 scanf ( "%lf%lf" , &point[i].x , &point[i].y ) ;
34 struct centre ctr = get_centre();
35 printf ( "%.2lf %.2lf\n" , ctr.x , ctr.y ) ;
36 }
37 return 0 ;
38 }