poj 2007 Scrambled Polygon(极角排序)
http://poj.org/problem?id=2007
Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the vertices of the polygon. When one starts at any vertex of a closed polygon and traverses each bounding line segment exactly once, one comes back to the starting vertex.
A closed polygon is called convex if the line segment joining any two points of the polygon lies in the polygon. Figure 1 shows a closed polygon which is convex and one which is not convex. (Informally, a closed polygon is convex if its border doesn't have any "dents".) The first property is that the vertices of the polygon will be confined to three or fewer of the four quadrants of the coordinate plane. In the example shown in Figure 2, none of the vertices are in the second quadrant (where x < 0, y > 0). To describe the second property, suppose you "take a trip" around the polygon: start at (0, 0), visit all other vertices exactly once, and arrive at (0, 0). As you visit each vertex (other than (0, 0)), draw the diagonal that connects the current vertex with (0, 0), and calculate the slope of this diagonal. Then, within each quadrant, the slopes of these diagonals will form a decreasing or increasing sequence of numbers, i.e., they will be sorted. Figure 3 illustrates this point. Input The input lists the vertices of a closed convex polygon in the plane. The number of lines in the input will be at least three but no more than 50. Each line contains the x and y coordinates of one vertex. Each x and y coordinate is an integer in the range -999..999. The vertex on the first line of the input file will be the origin, i.e., x = 0 and y = 0. Otherwise, the vertices may be in a scrambled order. Except for the origin, no vertex will be on the x-axis or the y-axis. No three vertices are colinear.
Output The output lists the vertices of the given polygon, one vertex per line. Each vertex from the input appears exactly once in the output. The origin (0,0) is the vertex on the first line of the output. The order of vertices in the output will determine a trip taken along the polygon's border, in the counterclockwise direction. The output format for each vertex is (x,y) as shown below.
Sample Input 0 0 Sample Output (0,0) Source |
××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××××
这题,,,,,这什么玩意啊
推荐个网站:http://www.cnblogs.com/devtang/archive/2012/02/01/2334977.html
《叉积排序,也就是可以排180度以内的,超出就会出错,
因为正弦函数在180内为正数,180到360为负数。》
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <math.h> #define MAXX 105 using namespace std; typedef struct point
{
int x,y;
}point;
typedef struct line
{
point st,ed;
}beline; int crossProduct(point a,point b,point c)
{
return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
} double Dist(point a,point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
} point c[MAXX];
point stk[MAXX];
int top;
bool cmp(point a,point b)
{
int len=crossProduct(c[],a,b);
if(len == )
return Dist(c[],a)<Dist(c[],b);
else
return len<;
} int main()
{
int i,j,k,t,x,y;
i=;
while(scanf("%d%d",&x,&y)!=EOF)
{
c[i].x=x;
c[i].y=y;
i++;
}
sort(c+,c+i,cmp);
for(int j=; j<i; j++)
printf("(%d,%d)\n",c[j].x,c[j].y);
}
poj 2007 Scrambled Polygon(极角排序)的更多相关文章
- poj 2007 Scrambled Polygon 极角排序
/** 极角排序输出,,, 主要atan2(y,x) 容易失精度,,用 bool cmp(point a,point b){ 5 if(cross(a-tmp,b-tmp)>0) 6 retur ...
- POJ 2007 Scrambled Polygon 极角序 水
LINK 题意:给出一个简单多边形,按极角序输出其坐标. 思路:水题.对任意两点求叉积正负判断相对位置,为0则按长度排序 /** @Date : 2017-07-13 16:46:17 * @File ...
- POJ 2007 Scrambled Polygon [凸包 极角排序]
Scrambled Polygon Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 8636 Accepted: 4105 ...
- 简单几何(极角排序) POJ 2007 Scrambled Polygon
题目传送门 题意:裸的对原点的极角排序,凸包貌似不行. /************************************************ * Author :Running_Time ...
- POJ 2007 Scrambled Polygon (简单极角排序)
题目链接 题意 : 对输入的点极角排序 思路 : 极角排序方法 #include <iostream> #include <cmath> #include <stdio. ...
- POJ 2007 Scrambled Polygon(简单极角排序)
水题,根本不用凸包,就是一简单的极角排序. 叉乘<0,逆时针. #include <iostream> #include <cstdio> #include <cs ...
- ●POJ 2007 Scrambled Polygon
题链: http://poj.org/problem?id=2007 题解: 计算几何,极角排序 按样例来说,应该就是要把凸包上的i点按 第三像限-第四像限-第一像限-第二像限 的顺序输出. 按 叉积 ...
- POJ 2007 Scrambled Polygon 凸包
Scrambled Polygon Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 7214 Accepted: 3445 ...
- POJ 2007 Scrambled Polygon 凸包点排序逆时针输出
题意:如题 用Graham,直接就能得到逆时针的凸包,找到原点输出就行了,赤果果的水题- 代码: /* * Author: illuz <iilluzen[at]gmail.com> * ...
随机推荐
- DirectoryInfo类
DirectoryInfo类和Directory类之间的关系与FileInfo类和File类之间的关系十分类似.下面介绍一下DirectoryInfo类的常用属性. DirectoryInfo类的常用 ...
- pstack使用和原理【转】
转自:http://www.cnblogs.com/mumuxinfei/p/4366708.html 前言: 最近小组在组织<<深入剖析Nginx>>的读书会, 里面作者提到 ...
- 【jQuery UI 1.8 The User Interface Library for jQuery】.学习笔记.1.CSS框架和其他功能
jquery.ui.all.css 1.所有主题必须的文件都包含在这个文件中.它由ui.base.css和ui.them.css两个文件中拉入的@import执行构成. jquery.ui.base. ...
- POJ 2192 :Zipper(DP)
http://poj.org/problem?id=2192 Zipper Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 1 ...
- 【转】SVN建库方法
转载地址:http://blog.csdn.net/winonatong/article/details/5791919 SVN全名Subversion,即版本控制系统.SVN与CVS一样,是一个跨平 ...
- poj2482 Stars in Your Window
此题可用线段树或静态二叉树来做. 考虑用线段树: 很容易想到先限定矩形横轴范围再考虑在此纵轴上矩形内物品总价值的最大值. 那么枚举矩形横轴的复杂度是O(n)的,考虑如何快速获取纵轴上的最大值. 我们不 ...
- 周赛-kiki's game 分类: 比赛 2015-08-02 09:24 7人阅读 评论(0) 收藏
kiki's game Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 40000/10000 K (Java/Others) Total S ...
- 浅谈算法和数据结构: 七 二叉查找树 八 平衡查找树之2-3树 九 平衡查找树之红黑树 十 平衡查找树之B树
http://www.cnblogs.com/yangecnu/p/Introduce-Binary-Search-Tree.html 前文介绍了符号表的两种实现,无序链表和有序数组,无序链表在插入的 ...
- Poj(1459),最大流,EK算法
题目链接:http://poj.org/problem?id=1459 Power Network Time Limit: 2000MS Memory Limit: 32768K Total Su ...
- 改变CSS样式
改变CSS样式 1.改变HTML元素样式的语法 //改变HTML样式的语法 document.getElementById(id).style.property = new style 例子: < ...