POJ 2194 2850 计算几何】的更多相关文章

题意: 给你了n个圆,让你摞起来,问顶层圆心的坐标 (数据保证间隔两层的圆不会挨着) 思路: 按照题意模拟. 假设我们已经知道了一层两个相邻圆的坐标a:(x1,y1)和b:(x2,y2) 很容易求出来边长是2,2,dis(a,b)的三角形的面积 进而求出来底面所对应的高 找到底面中点 讲a->b 向量旋转90度  乘上高度 就搞出来了坐标 //By SiriusRen #include <cmath> #include <cstdio> #include <algori…
题目链接:POJ 1410 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9) end point: (11,2) rectangle: left-top: (1,5) right-bottom: (7,1) Figure 1: Line se…
题目链接 切计算几何,感觉计算几何的算法还不熟.此题,枚举线段和圆点的直线,平分一个圆 #include <iostream> #include <cstring> #include <cstdio> #include <cstdlib> #include <cmath> using namespace std; #define eps 1e-8 struct point { double x,y; }p[]; double dis(point…
链接:http://poj.org/problem?id=2507 题意:哪个直角三角形,一直角边重合, 斜边分别为 X, Y, 两斜边交点高为 C , 求重合的直角边长度~ 思路: 设两个三角形不重合的两条直角边长为 a , b,根据 三角形相似, 则有 1/a + 1/b =1/c, 二分枚举答案得之~ #include <cstdio> #include <cmath> #include <iostream> #include <algorithm>…
题目大意:给你一个矩形的左上角和右下角的坐标,然后这个矩形有 N 个隔板分割成 N+1 个区域,下面有 M 组坐标,求出来每个区域包含的坐标数.   分析:做的第一道计算几何题目....使用叉积判断方向,然后使用二分查询找到点所在的区域.   代码如下: ==========================================================================================================================…
#include<stdio.h> #include<string.h> #include<iostream> #include<math.h> using namespace std; ]={,,,,,,,-,-,-}; ]={,-,,,-,,,-,,}; ]; __int64 area,x,y,px,py; int main() { int sum,t,tmp,i; cin>>tmp; while(tmp--) { scanf("%…
Calculate the number of toys that land in each bin of a partitioned toy box. 计算每一个玩具箱里面玩具的数量 Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toy…
题目链接:POJ 2254 Description As a member of an ACM programming team you'll soon find yourself always traveling around the world: Zürich, Philadelphia, San José, Atlanta,... from 1999 on the Contest Finals even will be on a different continent each year,…
TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12015   Accepted: 5792 Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away w…
意甲冠军:给出的一些段的.问:能否找到一条直线,通过所有的行 思维:假设一条直线的存在,所以必须有该过两点的线,然后列举两点,然后推断是否存在与所有的行的交点可以是 代码: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; struct Point { double x, y; Point() {} Point(dou…
跨产品的利用率推断点线段向左或向右,然后你可以2分钟 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 5005; int n, m, x1, y1, x2, y2; struct Point { int x, y; Point() {} Point(int x, int y) { this->x = x; this-&g…
Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to build a beautiful brick wall with a perfec…
Area of Simple Polygons Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3412   Accepted: 1763 Description There are N, 1 <= N <= 1,000 rectangles in the 2-D xy-plane. The four sides of a rectangle are horizontal or vertical line segment…
根据正方形对角的两顶点求另外两个顶点公式: x2 = (x1+x3-y3+y1)/2; y2 = (x3-x1+y1+y3)/2; x4= (x1+x3+y3-y1)/2; y4 = (-x3+x1+y1+y3)/2; #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=1000+5; struct Node { int x,y; bool…
Area Time Limit: 1000MS Memory Limit: 10000K Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the fol…
//第一期 计算几何题的特点与做题要领: 1.大部分不会很难,少部分题目思路很巧妙 2.做计算几何题目,模板很重要,模板必须高度可靠. 3.要注意代码的组织,因为计算几何的题目很容易上两百行代码,里面大部分是模板.如果代码一片混乱,那么会严重影响做题正确率. 4.注意精度控制. 5.能用整数的地方尽量用整数,要想到扩大数据的方法(扩大一倍,或扩大sqrt2).因为整数不用考虑浮点误差,而且运算比浮点快. 一.点,线,面,形基本关系,点积叉积的理解 POJ 2318 TOYS(推荐) http:/…
  Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1458   Accepted: 759 Description Given a triangle field and a rope of a certain length (Figure-1), you are required to use the rope to enclose a region within the field and make the regio…
Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28462   Accepted: 9498 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…
转自:http://blog.csdn.net/tyger/article/details/4480029 计算几何题的特点与做题要领:1.大部分不会很难,少部分题目思路很巧妙2.做计算几何题目,模板很重要,模板必须高度可靠.3.要注意代码的组织,因为计算几何的题目很容易上两百行代码,里面大部分是模板.如果代码一片混乱,那么会严重影响做题正确率.4.注意精度控制.5.能用整数的地方尽量用整数,要想到扩大数据的方法(扩大一倍,或扩大sqrt2).因为整数不用考虑浮点误差,而且运算比浮点快. 一.点…
http://poj.org/problem?id=2031 题意 给出三维坐标系下的n个球体,求把它们联通的最小代价. 分析 最小生成树加上一点计算几何.建图,若两球体原本有接触,则边权为0:否则边权为它们球心的距离-两者半径之和.这样来跑Prim就ok了.注意精度. #include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #in…
rt,计算几何入门: TOYS Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toy…
题目传送门:poj 2398 Toy Storage 题目大意:一个长方形的箱子,里面有一些隔板,每一个隔板都可以纵切这个箱子.隔板将这个箱子分成了一些隔间.向其中扔一些玩具,每个玩具有一个坐标,求有\(t​\)个玩具的隔间数(对\(t>0​\)都要输出). 题目分析:涉及到计算几何的知识是求点在线的哪一侧.可以利用叉积来做.取点\(A\)到隔板的上端点\(B\)的向量\(\vec{AB}\)叉乘点\(A\)到隔板的下端点\(C\)的向量\(\vec{AC}\).叉积的公式\(\vec a\ti…
题意: 给一些多边形或线段,输出与每一个多边形或线段的有哪一些多边形或线段. 解法: 想法不难,直接暴力将所有的图形处理成线段,然后暴力枚举,相交就加入其vector就行了.主要是代码有点麻烦,一步一步来吧. 还有收集了一个线段旋转的函数. 给定正方形对角求其他两点用到了线段旋转. Vector Rotate(Point P,Vector A,double rad){ //以P为基准点把向量A旋转rad return Vector(P.x+A.x*cos(rad)-A.y*sin(rad),P.…
题意: 在墙上钉两块木板,问能装多少水.即两条线段所夹的中间开口向上的面积(到短板的水平线截止) 解法: 如图: 先看是否相交,不相交肯定不行,然后就要求出P与A,B / C,D中谁形成的向量是指向上方的. 然后求出y值比较小的,建一条水平线,求出与另一条的交点,然后求面积. 要注意的是: 这种情况是不能装水的,要判掉. 还有 交G++会WA, 交C++就可以了, 不知道是POJ的问题还是 G++/C++的问题. 代码: #include <iostream> #include <cst…
Wall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28157   Accepted: 9401 Description Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he wo…
题意:给出一个100*100的正方形区域,通过若干连接区域边界的线段将正方形区域分割为多个不规则多边形小区域,然后给出宝藏位置,要求从区域外部开辟到宝藏所在位置的一条路径,使得开辟路径所需要打通的墙壁数最少("打通一堵墙"即在墙壁所在线段中间位置开一空间以连通外界),输出应打通墙壁的个数(包括边界上墙壁).题解:枚举每一个入口,在所有的情况中取穿墙数最少的输出即可,枚举每一个入口的时候,并不用枚举每条边的中间点,直接枚举该线段的两个顶点就行(因为要经过一个墙,那么从线段的任意地方进去都…
题目大意:有一个矩形盒子,盒子里会有一些木块线段,并且这些线段是按照顺序给出的,有n条线段,把盒子分层了n+1个区域,然后有m个玩具,这m个玩具的坐标是已知的,问最后每个区域有多少个玩具 解题思路:因为线段是有序给出,所以不用排序,判断某个点在哪个区域,采用二分法,将某个点和线段的叉积来判断这个点是在线的左边或者右边,根据这个来二分找出区域 代码和思路参考与此链接:http://blog.csdn.net/wangjian8006 这也算我入门计算几何的第一道题了,首先看了一些有关资料,了解到叉…
题目链接:http://poj.org/problem?id=1385 题目大意:给你一个多边形的点,求重心. 首先,三角形的重心: ( (x1+x2+x3)/3 , (y1+y2+y3)/3 ) 然后多边形的重心就是将多边形划分成很多个三角形,以三角形面积为权值,将每个三角形的重心加权平均. 注意: pair<double,double>会MLE.. fabs会损失精度?(这个我也不知道),因此在用向量叉积求三角形面积的时候最好是直接让面积求出来就是正的..否则fabs就WA了... 代码:…
Toy Storage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3146   Accepted: 1798 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…
poj 3968 (bzoj 2642) 二分+半平面交,每次不用排序,这是几个算几版综合. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<vector> #include<deque> using namespace std; #define MAXN 100000 na…