zoukankan      html  css  js  c++  java
  • Lifting the Stone(hdu1115)多边形的重心

    Lifting the Stone

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 5370 Accepted Submission(s): 2239


    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
     
     
    题意:求多边形的重心,,,,
     
    模版题。
    #include<stdio.h>
    #include<stdlib.h>
    /*==================================================*
    | 求多边形重心
    | INIT: pnt[]已按顺时针(或逆时针)排好序;
    | CALL: res = bcenter(pnt, n);
    *==================================================*/
    struct point
    {
        double x, y;
    }pnt[1000005],res;
    point bcenter(point pnt[], int n)//重心
    {
        point p, s;
        double tp, area = 0, tpx = 0, tpy = 0;
        p.x = pnt[0].x;
        p.y = pnt[0].y;
        for (int i = 1; i <= n; ++i)
        {
            // point: 0 ~ n-1
            s.x = pnt[(i == n) ? 0 : i].x;
            s.y = pnt[(i == n) ? 0 : i].y;
            tp = (p.x * s.y - s.x * p.y);//叉乘
            area += tp / 2;
            tpx += (p.x + s.x) * tp;
            tpy += (p.y + s.y) * tp;
            p.x = s.x;
            p.y = s.y;
        }
        s.x = tpx / (6 * area);
        s.y = tpy / (6 * area);
        return s;
    }
    int main()
    {
        int T,N,i;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d",&N);
            for(i=0;i<N;i++)
            {
                scanf("%lf%lf",&pnt[i].x,&pnt[i].y);
            }
            res=bcenter(pnt, N);
            printf("%0.2lf %0.2lf
    ",res.x,res.y);
        }
        return 0;
    }
     
  • 相关阅读:
    意外发现在调用Activator.CreateInstance的时候在构造函数处加断点居然可以~~
    手机操作系统
    读取Excel文件到DataSet
    支持mrp软件的手机(MTK手机)检测
    如何查看手机系统版本
    .NET进度条用例
    dos命令导出指定类型的文件列表
    FTP上传下载 FTP操作类 FTPHelper 异步上传 递归创建文件文件夹
    批量删除GridView(DataGrid)选中项
    sql判断临时表是否存在
  • 原文地址:https://www.cnblogs.com/yuyixingkong/p/3912433.html
Copyright © 2011-2022 走看看