Scrambled Polygon
题目：http://poj.org/problem?id=2007
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 6513 Accepted: 3092
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 subject of this problem is a closed convex polygon in the coordinate plane, one of whose vertices is the origin (x = 0, y = 0). Figure 2 shows an example. Such a polygon will have two properties significant for this problem.
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
70 -50
60 30
-30 -50
80 20
50 -60
90 -20
-30 -40
-10 -60
90 10
Sample Output
(0,0)
(-30,-40)
(-30,-50)
(-10,-60)
(50,-60)
(70,-50)
(90,-20)
(90,10)
(80,20)
(60,30)
计算几何，给出一系列点，按照极角排序。
应该属于Graham求凸包的前置题目吧。。。
很基础。注意输入时判定停止。
在注意一下精度，应该就OK了，
说一下极角排序几种方法吧，这些都是看人家Blog。
转自：http://www.cnblogs.com/devtang/archive/2012/02/01/2334977.html
1.利用叉积的正负来作cmp.(即是按逆时针排序).此题就是用这种方法
double cross(point p0,point p1,point p2)
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
// sort排序函数
bool cmp(const point &a, const point &b)//逆时针排序
{
point origin;
// 设置原点
origin.x = origin.y = 0;
return cross(origin,b,a)<EPS;
}
2.利用complex的内建函数。
#include<complex>
#define x real()
#define y imag()
#include<algorithm>
using namespace std;
bool cmp(const Point& p1, const Point& p2)
{
return arg(p1) < arg(p2);
}
3.利用arctan计算极角大小。（范围『-180，180』）
bool cmp(const Point& p1, const Point& p2)
{
return atan2(p1.y, p1.x) < atan2(p2.y, p2.x);
}
4.利用象限加上极角，叉积。
bool cmp(const point &a, const point &b)//先按象限排序，再按极角排序，再按远近排序
{
if (a.y == 0 && b.y == 0 && a.x*b.x <= 0)return a.x>b.x;
if (a.y == 0 && a.x >= 0 && b.y != 0)return true;
if (b.y == 0 && b.x >= 0 && a.y != 0)return false;
if (b.y*a.y <= 0)return a.y>b.y;
point one;
one.y = one.x = 0;
return cross(one,a,one,b) > 0 || (cross(one,a,one,b) == 0 && a.x < b.x);
}
恩，就是这些啦~
#include <stdio.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define EPS 1e-8
struct point
{
double x,y;
}p[51];
double cross(point p0,point p1,point p2)
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
// sort排序函数
bool cmp(const point &a, const point &b)//逆时针排序
{
point origin;
// 设置原点
origin.x = origin.y = 0;
return cross(origin,b,a)<EPS;
}
int main()
{
int i,n;
n=0;
while( scanf("%lf%lf",&p[n].x,&p[n].y)!=EOF )
{
++n;
}
// 极角排序，第一个空过去
sort(p+1,p+n,cmp);
for(i=0;i<n;++i)
printf("(%.0lf,%.0lf)\n",p[i].x,p[i].y);
return 0;
}