Hunter’s Apprentice 【判断多边形边界曲线顺逆时针】
问题 H: Hunter’s Apprentice
时间限制: 1 Sec 内存限制: 128 MB
提交: 353 解决: 39
[提交] [状态] [命题人:admin]
题目描述
When you were five years old, you watched in horror as a spiked devil murdered your parents. You would have died too, except you were saved by Rose, a passing demon hunter. She ended up adopting you and training you as her apprentice.
Rose’s current quarry is a clock devil which has been wreaking havoc on the otherwise quiet and unassuming town of Innsmouth. It comes out each night to damage goods, deface signs, and kill anyone foolish enough to wander around too late. The clock devil has offed the last demon hunter after it; due to its time-warping powers, it is incredibly agile and fares well in straight-up fights.
The two of you have spent weeks searching through dusty tomes for a way to defeat this evil. Eventually, you stumbled upon a relevant passage. It detailed how a priest managed to ensnare a clock devil by constructing a trap from silver, lavender, pewter, and clockwork. The finished trap contained several pieces, which must be placed one-by-one in the shape of a particular polygon, in counter-clockwise order. The book stated that the counter-clockwise order was important to counter the clock devil’s ability to speed its own time up, and that a clockwise order would only serve to enhance its speed.
It was your responsibility to build and deploy the trap, while Rose prepared for the ensuing fight. You carefully reconstruct each piece as well as you can from the book. Unfortunately, things did not go as planned that night. Before you can finish preparing the trap, the clock devil finds the two of you. Rose tries to fight the devil, but is losing quickly. However, she is buying you the time to finish the trap. You quickly walk around them in the shape of the polygon, placing each piece in the correct position. You hurriedly activate the trap as Rose is knocked out. Just then, you remember the book’s warning. What should you do next?
输入
The first line of input contains a single integer T (1 ≤ T ≤ 100), the number of test cases. The first line of each test case contains a single integer n (3 ≤ n ≤ 20), the number of pieces in the trap. Each of the next n lines contains two integers xi and yi (|xi |, |yi | ≤ 100), denoting the x and y coordinates of where the ith piece was placed. It is guaranteed that the polygon is simple; edges only intersect at vertices, exactly two edges meet at each vertex, and all vertices are distinct.
输出
For each test case, output a single line containing either fight if the trap was made correctly or run if the trap was made incorrectly.
题意:
你要去帮罗斯去打怪兽balabala,然后你得绕着一个多边形跑,逆时针能帮上忙,顺时针会帮倒忙
给你一堆点坐标,然后按点坐标顺序跑,判断是顺时针还是逆时针
这有个公式,如果知道公式就是简单题了,
如果不知道又像我这种一辈子都不可能手推出来公式的就只能表示 模板+1
关于公式 https://blog.csdn.net/henuyh/article/details/80378818
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define rep(i,a,n) for(int i=a;i<n;++i)
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define sca(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define mst(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
typedef pair<int,int> P;
typedef long long ll;
const int INF =0x3f3f3f3f;
const int inf =0x3f3f3f3f;
const int mod = 1e9+7;
const int MAXN = 105;
const int maxn = 10010;
double x[1050],y[1050];
int main()
{
int t;
read(t);
while(t--)
{
int n;
read(n);
double temp = 0;
rep(i, 0, n)
{
scanf("%lf%lf", &x[i], &y[i]);
}
rep(i,0,n-1)
{
temp += -0.5 * (y[i + 1] + y[i]) * (x[i + 1] - x[i]); //Green 公式
}
if(temp > 0)
cout<<"fight"<<endl;
else
cout<<"run"<<endl;
}
return 0;
}
Hunter’s Apprentice 【判断多边形边界曲线顺逆时针】的更多相关文章
- upc组队赛5 Hunter’s Apprentice 【判断多边形边界曲线顺逆时针】
Hunter's Apprentice 题目描述 When you were five years old, you watched in horror as a spiked devil murde ...
- 【ECNU3386】Hunter's Apprentice(多边形有向面积)
点此看题面 大致题意: 按顺序给你一个多边形的全部顶点,让你判断是按顺时针还是逆时针给出的. 多边形有向面积 显然我们知道,叉积可以求出两个向量之间有向面积的两倍. 所以,我们三角剖分,就可以求出四边 ...
- Hunter’s Apprentice(判断所走路线为顺时针或逆时针)【Green公式】
Hunter's Apprentice 题目链接(点击) 题目描述 When you were five years old, you watched in horror as a spiked de ...
- POJ1584 判断多边形是否为凸多边形,并判断点到直线的距离
求点到直线的距离: double dis(point p1,point p2){ if(fabs(p1.x-p2.x)<exp)//相等的 { return fabs(p2.x-pe ...
- hdu 2108:Shape of HDU(计算几何,判断多边形是否是凸多边形,水题)
Shape of HDU Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Tota ...
- hdu 2108 Shape of HDU【判断多边形是否是凸多边形模板】
链接: http://acm.hdu.edu.cn/showproblem.php?pid=2108 http://acm.hust.edu.cn/vjudge/contest/view.action ...
- poj1584 A Round Peg in a Ground Hole 判断多边形凹凸,点到线的距离【基础计算几何】
大致思路:首先对于所给的洞的点,判断是否是凸多边形,图形的输入和输出可以是顺时针或者逆时针,而且允许多点共线 Debug 了好几个小时,发现如下问题 判断三点是否共线,可用斜率公式判断 POINT p ...
- hdu2108 Shape of HDU 极角排序判断多边形
Problem Description 话说上回讲到海东集团推选老总的事情,最终的结果是XHD以微弱优势当选,从此以后,"徐队"的称呼逐渐被"徐总"所取代,海东 ...
- TZOJ 2560 Geometric Shapes(判断多边形是否相交)
描述 While creating a customer logo, ACM uses graphical utilities to draw a picture that can later be ...
随机推荐
- Android -- 实现RecyclerView可拖拽Item
1,今天和大家一起实现RecyclerView可拖拽Item,主要是使用RecyclerView结合ItemTouchHelper来实现的,来看一下效果 2,看一下怎么实现的呢,很简单,只需要给rec ...
- Python全栈-网络编程基础
一.C/S架构 1.硬件C/S架构 如PC-打印机 2.软件C/S架构 如PC-网站服务器 参照: https://baike.baidu.com/item/Client%2FServer/15044 ...
- [openjudge-动态规划]怪盗基德的滑翔翼
题目描述 描述 怪盗基德是一个充满传奇色彩的怪盗,专门以珠宝为目标的超级盗窃犯.而他最为突出的地方,就是他每次都能逃脱中村警部的重重围堵,而这也很大程度上是多亏了他随身携带的便于操作的滑翔翼. 有一天 ...
- arm cortex-m0plus源码学习(二)AMBA3.0_ AHBLite
1. AMBA总线概述 AMBA2.0 以上版本都是基于单沿时钟.单向信号线的协议[1]. 现在市场上大部分的基于 AMBA 架构的 SoC 产品, 系统总线采用 AHB, 外部总线采用 APB.系统 ...
- Python使用闭包结合配置自动生成函数
背景 在构建测试用例集时,常常需要编写一些函数,这些函数接受基本相同的参数,仅有一个参数有所差异,并且处理模式也非常相同.可以使用Python闭包来定义模板函数,然后通过参数调节来自动化生产不同的函数 ...
- 解读 JavaScript 之引擎、运行时和堆栈调用
https://www.oschina.net/translate/how-does-javascript-actually-work-part-1 随着 JavaScript 变得越来越流行,很多团 ...
- AWS免费云服务套餐申请步骤及常见问题
AWS免费云服务套餐申请步骤及常见问题 AWS免费使用套餐常见问题_AWS免费云服务套餐_-AWS云服务https://amazonaws-china.com/cn/free/faqs/ 什么是 AW ...
- 如何干净卸载mysql
一.在控制面板中卸载mysql软件: 二.卸载过后删除C:\Program Files (x86)\MySQL该目录下剩余了所有文件,把mysql文件夹也删了: 三.windows+R运行“reged ...
- [转载]oracle的加密和解密
加密函数 create or replace function encrypt_des(p_text varchar2, p_key varchar2) return varchar2 isv_tex ...
- 自学Java第二周的总结
在这一周里我在网上学习了java的对象和类,了解了对象与类以及简单的用法.对象是类的一个实例(对象不是找个女朋友),有状态和行为.例如,一条狗是一个对象,它的状态有:颜色.名字.品种:行为有:摇尾巴. ...