zoukankan      html  css  js  c++  java
  • POJ 1385 Lifting the Stone (多边形的重心)

    Lifting the Stone

    题目链接:

    http://acm.hust.edu.cn/vjudge/contest/130510#problem/G

    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

    2016-HUST-线下组队赛-5
    ##题意: 求多边形的重心.
    ##题解: 裸到爆的求重心,开场就打了,硬是最后才过掉. 坑在四舍五入时出现了-0.00的情况导致WA... 涨姿势了,以后切记.
    ##代码: ``` cpp #include #include #include #include #include #include #include #include #include #include #define maxn 1110000 #define LL long long #define eps 1e-8 #define inf 0x3f3f3f3f #define mod 1000000007 #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;

    int n;
    struct pt{
    double x,y;
    }p[maxn];

    double xmul(pt p0, pt p1, pt p2) {
    return (p1.x-p0.x)(p2.y-p0.y) - (p2.x-p0.x)(p1.y-p0.y);
    }

    pt center(int n) {
    pt ret = {0,0}, t;
    double t1 = 0, t2;
    for(int i=1; i<n-1; i++) {
    t2 = xmul(p[i+1],p[0],p[i]);
    t.x = (p[0].x + p[i].x + p[i+1].x) / 3.0;
    t.y = (p[0].y + p[i].y + p[i+1].y) / 3.0;
    ret.x += t.x * t2;
    ret.y += t.y * t2;
    t1 += t2;
    }

    ret.x /= t1, ret.y /= t1;
    return ret;
    

    }

    int main()
    {
    //IN;

    int T; cin >> T;
    while(T--)
    {
        scanf("%d", &n);
        for(int i=0; i<n; i++) {
            scanf("%lf %lf", &p[i].x, &p[i].y);
        }
    
        pt ans = center(n);
        if(ans.x > -0.005 && ans.x < 0.005) ans.x = 0.0;
        if(ans.y > -0.005 && ans.y < 0.005) ans.y = 0.0;
        printf("%.2f %.2f
    ", ans.x, ans.y);
    }
    
    return 0;
    

    }

  • 相关阅读:
    ztCreateUserWizard输入密码和设置安全问题
    寻觅在office(确切的说是word) 的工具栏中添加控件的方法,找到了这个控件列表
    vs2005,combox 数据绑定和SelectedIndexChanged事件触发 引发的问题
    今天下载安装了Enterprise Library for .NET Framework 2.0 January 2006,准备试试dataAccess application block
    zt 3DO的历史,让人感慨啊
    iBM,tivoli,flash挺有意思的
    zt科学家发现自转最快中子星 比地球快1亿倍
    c#中读取应用程序路径的方法
    郁闷啊郁闷,为了access的like,折腾了一上午.
    人生
  • 原文地址:https://www.cnblogs.com/Sunshine-tcf/p/5816456.html
Copyright © 2011-2022 走看看