Rotational Painting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2498    Accepted Submission(s): 702 Problem Description Josh Lyman is a gifted painter. One of his great works is a glass pain…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6971    Accepted Submission(s): 2919 Problem Description There are many secret openings in the floor which are covered by a big…

Problem Description Josh Lyman is a gifted painter. One of his great works is a glass painting. He creates some well-designed lines on one side of a thick and polygonal glass, and renders it by some special dyes. The most fantastic thing is that it c…

F - Rotational Painting Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3685 Appoint description:  System Crawler  (2014-11-09) Description Josh Lyman is a gifted painter. One of his great works…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4819    Accepted Submission(s): 2006 Problem Description There are many secret openings in the floor which are covered by a big…

题意:已知一多边形没有边相交,质量分布均匀.顺序给出多边形的顶点坐标,求其重心. 分析: 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.这个题就是这一类型,算法和上面的不同. 特殊地,质量均匀的三角形重心: X = ( x0 + x1 + x2 ) / 3…

题意是给一个 n 边形,给出沿逆时针方向分布的各顶点的坐标,求出 n 边形的重心. 求多边形重心的情况大致上有三种: 一.多边形的质量都分布在各顶点上,像是用轻杆连接成的多边形框,各顶点的坐标为Xi,Yi,质量为mi,则重心坐标为: X = ∑( xi * mi ) /  ∑ mi ; Y = ∑( yi * mi)  / ∑ mi; 若每个顶点的质量相等,则重心坐标为: X = ∑ xi / n; Y = ∑ yi / n; 二.多边形的质量分布均匀,像是用密度相同的材料制成的多边形板子,多采…

题意:求一个不规则简单多边形的重心. 解法:多边形的重心就是所有三角形的重心对面积的加权平均数. 关于求多边形重心的文章: 求多边形重心 用叉积搞一搞就行了. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #define Mod 100000000…

Lifting the Stone http://acm.hdu.edu.cn/showproblem.php?pid=1115 题目描述:输入n个顶点(整数),求它们围成的多边形的重心. 算法:以一个点出发,与其他非邻点相连,将n边形划分成n-2个三角形.求每个三角形的质点系重心(如:((x1+x2+x3)/3,(y1+y2+y3)/3)),再求出每个三角形的面积.相乘求和后除以多边形面积). 注意:we connect the points in the given order.输入的顺序,…

1.http://acm.hdu.edu.cn/showproblem.php?pid=2440   按照题意知道是一个简单的多边形即凸包,但给出的点并没有按照顺序的,所以需要自己先求出凸包,然后在用随机淬火求费马点. #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<cstdlib> #inc…