首页 > ACM题库 > HDU-杭电 > HDU 1115 Lifting the Stone-计算几何-[解题报告] C++
2013
11-27

HDU 1115 Lifting the Stone-计算几何-[解题报告] C++

Lifting the Stone

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

2
4
5 0
0 5
-5 0
0 -5
4
1 1
11 1
11 11
1 11

样例输出:

0.00 0.00
6.00 6.00

/*
摘自:KIDxの博客 
求多边形重心的题目大致有这么几种: 
1、质量集中在顶点上 
    n个顶点坐标为(xi,yi),质量为mi,则重心 
  X = ∑( xi×mi ) / ∑mi 
  Y = ∑( yi×mi ) / ∑mi 
  特殊地,若每个点的质量相同,则 
  X = ∑xi / n 
  Y = ∑yi / n 
2、质量分布均匀 
  特殊地,质量均匀的三角形重心: 
  X = ( x0 + x1 + x2 ) / 3 
  Y = ( y0 + y1 + y2 ) / 3 
3、质量分布不均匀 
    只能用函数多重积分来算,不太会 
这题的做法: 
将n边形分成多个三角形,分别求出重心坐标以及质量m【因为质量分布均匀,所以可以设密度为1,则面积就是质量】 
因为质量都集中在重心 
所以把所有求出来的重心按逆时针连接起来又是一个多边形 
但是这个多边形的质量集中在顶点上 
所以可以利用上面公式进行计算 
*/
#include <cstdio>
#include <cstdlib>
#include <iostream>

using namespace std;

const int N = 1000000;

struct point {
    double x;
    double y;
}p[N], g;

double crossProd(point A, point B, point C) {
    return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
}

void compGravity(int n) {//求重心 
    point tmp;
    double sumArea, area;
    sumArea = 0;
    g.x = g.y = 0;
    for (int i=2; i<n; ++i) {
        area = crossProd(p[0], p[i-1], p[i]);
        sumArea += area;
        tmp.x = p[0].x + p[i-1].x + p[i].x;
        tmp.y = p[0].y + p[i-1].y + p[i].y;
        g.x += tmp.x * area;
        g.y += tmp.y * area;
    }
    g.x /= (sumArea * 3.0);
    g.y /= (sumArea * 3.0);
}

int main() {
    int t;
    scanf ("%d", &t);
    while (t--) {
        int n;
        scanf ("%d", &n);//顶点以逆时针方向给出 
        for (int i=0; i<n; ++i) scanf ("%lf%lf", &p[i].x, &p[i].y);
        compGravity(n);
        printf ("%.2lf %.2lf\n", g.x, g.y);
    }
    return 0;
}

 


  1. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
    O(n) = O(n-1) + O(n-2) + ….
    O(n-1) = O(n-2) + O(n-3)+ …
    O(n) – O(n-1) = O(n-1)
    O(n) = 2O(n-1)

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