B - Scrambled Polygon

B - Scrambled Polygon

Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  POJ 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 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)

题目大意：对任意给出的一些点（保证围成的是凸边形）排序按逆时钟输出该突边形的点（看输出和图就会懂了）
借此题 又记忆了一遍qsort排序 恩 不错的感觉！
思路：这里大致说一下是用凸包算法（不懂可以网上查一下），凸包是经典的计算几何学问题，判断向量p1=(x1,y1)到p2=(x2,y2)是否做左

转，只需要判断x1*y2-x2*y1（这个又是叉积吧！！！） 的正负，如果结果为正，则从p1 到p2 做左转。哎，做这题时竟然不知到什么时候退出！！！不知道你有这种感觉没！

#include
using namespace std;
struct point
{
	int x;
	int y;
};
int cmp(const void *a,const void *b)   //排序调用的函数，用那个凸包原理
{
	struct point *p,*q;
	int i;
	p = ( struct point * )a;
	q = ( struct point * )b;
	i = p->x*q->y - p->y*q->x;
	if(i>0)	return -1;
	return 1;
}
int main()
{
	point a[102];
	int i=1,j;
	cin>>a[0].x>>a[0].y;
	while(cin>>a[i].x>>a[i].y)//不知道什么时候退出，自己测试时写成了while(cin>>a[i].x>>a[i].y&&a[i].x)
		i++;
	qsort(a+1,i-1,sizeof(a[0]),cmp);
	cout<<"(0,0)"<