UVA LIVE-3263 - That Nice Euler Circuit
画一个顶点为偶数的封闭的二维图,当然。这个图能够自交,给出画的过程中的一些轨迹点。求出这个图把二次元分成了几部分,比如三角形把二次元分成了两部分。
这个的话,有图中顶点数+部分数-棱数=2的定律,这是核心思想。也就是所谓的欧拉定律拓扑版,好吧,事实上细致想想也是可以想出这个规律来的。
做出这题纯属意外,因为给的点的坐标全是用整数表示,为了不用考虑精度问题,一開始。我就想仅仅用这些点。就是说不再算出其他交点之类的,就把答案算出,
由于当前轨迹与之前轨迹无非三种情况:规范与不规范相交,不相交
不相交当然就不用管了,相交的话,考虑两种情况下的顶点、棱的数量变化
可是不知道是不是题意有些细节没理解到,还是有些特殊的图的情况没考虑到。用这样的思路尽管代码比較精炼。可是一直wa
后来索性干脆把全部的顶点用行列式求出,然后再求棱。略微想想能够知道,棱的数目是在轨迹数目的基础上加上在轨迹中间,即在轨迹上。不在轨迹两点的交点数目
这里就须要通过叉积推断点是否在轨迹中间。因为题中没有给出精度要求……假设不给一个精度,而直接用叉积为0推断点是否在轨迹所在的直线上的话,会wa
由于直接用0。相当于。精度就是double的精度,也就是1e-15,所以后来改成了1e-9然后就过了。
须要注意的是,在杭电上也有相同的题,但明显杭电的oj比UVA的渣,杭电的g++比c++的精度运算损失少,所以在杭电上交用这样的不太好的方法写的代码,须要用g++交才干
过。
我的代码:
#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const double inf=1e4,eps=1e-9;
struct dot
{
double x,y;
dot(){}
dot(double a,double b){x=a;y=b;}
dot operator -(dot a){return dot(x-a.x,y-a.y);}
friend bool operator <(dot a,dot b){return a.x!=b.x?a.x<b.x:a.y<b.y;}
bool operator ==(dot a){return x==a.x&&y==a.y;}
bool operator !=(dot a){return x!=a.x||y!=a.y;}
double operator *(dot a){return x*a.y-y*a.x;}
double dis(dot a){return sqrt(pow(x-a.x,2)+pow(y-a.y,2));}
};
bool isdil(dot a,dot b,dot c)
{
return a.x<=max(b.x,c.x)&&
a.y<=max(b.y,c.y)&&
min(b.x,c.x)<=a.x&&
min(b.y,c.y)<=a.y&&
a!=b&&a!=c;
}
bool isdbl(dot a,dot b,dot c){return fabs((a-c)*(b-c))<eps&&isdil(a,b,c);}
dot cross(dot a,dot b,dot c,dot d)
{
double e,f,g,h,i,j,k,l,m;
e=b.y-a.y;f=a.x-b.x;g=a.x*b.y-a.y*b.x;
h=d.y-c.y;i=c.x-d.x;j=c.x*d.y-c.y*d.x;
k=dot(e,h)*dot(f,i);
if(k==0)
return dot(inf,inf);
l=dot(g,j)*dot(f,i);
m=dot(e,h)*dot(g,j);
dot t=dot(l/k,m/k);
return isdil(t,a,b)&&isdil(t,c,d)?t:dot(inf,inf);
}
int main()
{
dot a[310],b[30000],t;
int i,n,j,k,ans,T=0;
while(cin>>n&&n)
{
for(i=0;i<n;i++)
{
cin>>a[i].x>>a[i].y;
b[i]=a[i];
}
k=n;
for(i=1;i<n;i++)
for(j=i+2;j<n;j++)
{
t=cross(a[i-1],a[i],b[j-1],b[j]);
if(t.x!=inf)
b[k++]=t;
}
sort(b,b+k);
k=unique(b,b+k)-b;
ans=1+n-k;
for(j=0;j<k;j++)
for(i=1;i<n;i++)
if(isdbl(b[j],a[i-1],a[i]))
ans++;
printf("Case %d: There are %d pieces.\n",++T,ans);
}
}
原题:
Description

Little Joey invented a scrabble machine that he called Euler, after the great mathematician. In his primary school Joey heard about the nice story of how Euler started the study about graphs. The problem in that story was - let me remind you - to draw a
graph on a paper without lifting your pen, and finally return to the original position. Euler proved that you could do this if and only if the (planar) graph you created has the following two properties: (1) The graph is connected; and (2) Every vertex in
the graph has even degree.
Joey's Euler machine works exactly like this. The device consists of a pencil touching the paper, and a control center issuing a sequence of instructions. The paper can be viewed as the infinite two-dimensional plane; that means you do not need to worry about
if the pencil will ever go off the boundary.
In the beginning, the Euler machine will issue an instruction of the form
(X0, Y0) which moves the pencil to some starting position(X0,Y0). Each subsequent instruction is also of the form(X',Y'), which means to move the
pencil from the previous position to the new position(X',Y'), thus draw a line segment on the paper. You can be sure that the new position is different from the previous position for each instruction. At last, the
Euler machine will always issue an instruction that move the pencil back to the starting position(X0,
Y0). In addition, the Euler machine will definitely not draw any lines that overlay other lines already drawn. However, the lines may intersect.
After all the instructions are issued, there will be a nice picture on Joey's paper. You see, since the pencil is never lifted from the paper, the picture can be viewed as an Euler circuit.
Your job is to count how many pieces (connected areas) are created on the paper by those lines drawn by Euler.
Input
There are no more than 25 test cases. Ease case starts with a line containing an integerN4,
which is the number of instructions in the test case. The followingN pairs of integers give the instructions and appear on a single line separated by single spaces. The first pair is the first instruction that gives the coordinates
of the starting position. You may assume there are no more than 300 instructions in each test case, and all the integer coordinates are in the range (-300, 300). The input is terminated whenN is 0.
Output
For each test case there will be one output line in the format
Case x: There are w pieces.,
where x is the serial number starting from 1.
Note: The figures below illustrate the two sample input cases.

Sample Input
5
0 0 0 1 1 1 1 0 0 0
7
1 1 1 5 2 1 2 5 5 1 3 5 1 1
0
Sample Output
Case 1: There are 2 pieces.
Case 2: There are 5 pieces.
UVA LIVE-3263 - That Nice Euler Circuit的更多相关文章
- UVALive - 3263 That Nice Euler Circuit (几何)
UVALive - 3263 That Nice Euler Circuit (几何) ACM 题目地址: UVALive - 3263 That Nice Euler Circuit 题意: 给 ...
- UVALi 3263 That Nice Euler Circuit(几何)
That Nice Euler Circuit [题目链接]That Nice Euler Circuit [题目类型]几何 &题解: 蓝书P260 要用欧拉定理:V+F=E+2 V是顶点数; ...
- LA 3263 That Nice Euler Circuit(欧拉定理)
That Nice Euler Circuit Little Joey invented a scrabble machine that he called Euler, after the grea ...
- 简单几何(求划分区域) LA 3263 That Nice Euler Circuit
题目传送门 题意:一笔画,问该图形将平面分成多少个区域 分析:训练指南P260,欧拉定理:平面图定点数V,边数E,面数F,则V + F - E = 2.那么找出新增的点和边就可以了.用到了判断线段相 ...
- uvalive 3263 That Nice Euler Circuit
题意:平面上有一个包含n个端点的一笔画,第n个端点总是和第一个端点重合,因此团史一条闭合曲线.组成一笔画的线段可以相交,但是不会部分重叠.求这些线段将平面分成多少部分(包括封闭区域和无限大区域). 分 ...
- UVAlive 3263 That Nice Euler Circuit(欧拉定理)
题目链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=21363 [思路] 欧拉定理:V+F-E=2.则F=E-V+2. 其 ...
- UVALive 3263: That Nice Euler Circuit (计算几何)
题目链接 lrj训练指南 P260 //==================================================================== // 此题只需要考虑线 ...
- UVa 10735 - Euler Circuit(最大流 + 欧拉回路)
链接: https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem& ...
- UVa 10735 (混合图的欧拉回路) Euler Circuit
题意: 给出一个图,有的边是有向边,有的是无向边.试找出一条欧拉回路. 分析: 按照往常的思维,遇到混合图,我们一般会把无向边拆成两条方向相反的有向边. 但是在这里却行不通了,因为拆成两条有向边的话, ...
随机推荐
- 2.SpringBoot之返回json数据
一.创建一个springBoot个项目 操作详情参考:1.SpringBoo之Helloword 快速搭建一个web项目 二.编写实体类 /** * Created by CR7 on 2017-8- ...
- Java编程思想第四版第二章练习题答案
练习1:创建一个类,它包含一个int域和一个char域,它们都没有被初始化.将他们的值打印出来,以验证Java执行了默认初始化 public class JavaThinking { private ...
- 2016-2017-2 20155309 南皓芯《java程序设计》第八周学习总结
教材学习内容总结 本周学习的主要是第十四章,第十五章的内容. NIO与NIO2 同步非阻塞IO(Java NIO) : 同步非阻塞,服务器实现模式为一个请求一个线程,即客户端发送的连接请求都会注册到多 ...
- Redis(四)Redis高级
一Redis 数据备份与恢复 Redis SAVE 命令用于创建当前数据库的备份. 语法 redis Save 命令基本语法如下: redis 127.0.0.1:6379> SAVE 实例 r ...
- day6面向对象
面向对象介绍(http://www.cnblogs.com/alex3714/articles/5188179.htm) 世界万物,皆可分类 世界万物,皆为对象 只要是对象,就 ...
- USACO 6.5 All Latin Squares
All Latin Squares A square arrangement of numbers 1 2 3 4 5 2 1 4 5 3 3 4 5 1 2 4 5 2 3 1 5 3 1 2 4 ...
- 跟我一起学WPF(0):初识WPF
WPF是什么 WPF是微软的新一代图形引擎系统,全称为Windows Presentation Foundation,从.NET3.0版本开始引入,为用户界面.2D/3D 图形.文档和媒体提供了统一的 ...
- 微信公众号开发--用.Net Core实现微信消息加解密
1.进入微信公众号后台设置微信服务器配置参数(注意:Token和EncodingAESKey必须和微信服务器验证参数保持一致,不然验证不会通过). 2.设置为安全模式 3.代码实现(主要分为验证接口和 ...
- poj1730 - Perfect Pth Powers(完全平方数)(水题)
/* 以前做的一道水题,再做精度控制又出了错///... */ 题目大意: 求最大完全平方数,一个数b(不超过int范围),n=b^p,使得给定n,p最大: 题目给你一个数n,求p : 解题思路: 不 ...
- nc工具学习
0x00.命令详解 基本使用 想要连接到某处:nc [-options] ip port 绑定端口等待连接:nc -l -p port ip 参数: -e prog 程序重定向,一旦连接,就执行 [ ...