You can Solve a Geometry Problem too

最新推荐文章于 2024-08-02 10:00:00 发布

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#output #each #integer #input #float

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本文深入探讨了ACM/ICPC中常见几何问题，特别聚焦于线段相交计数的简化算法。通过实例演示，提供了解决几何问题的直观方法，并强调了简化步骤的重要性。

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http://acm.hdu.edu.cn/showproblem.php?pid=1086

Problem Description

Many geometry（几何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :)
Give you N (1<=N<=100) segments（线段）, please output the number of all intersections（交点）. You should count repeatedly if M (M>2) segments intersect at the same point.

Note:
You can assume that two segments would not intersect at more than one point.

Input

Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending.
A test case starting with 0 terminates the input and this test case is not to be processed.

Output

For each case, print the number of intersections, and one line one case.

Sample Input

  
   2
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.00
3
0.00 0.00 1.00 1.00
0.00 1.00 1.00 0.000
0.00 0.00 1.00 0.00
0

Sample Output

1
3

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
求线段相交，交点个数；不过不是很明白为什么不用判平行。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
struct node
{
	double x1,x2,y1,y2;
}g[150];
int cmp(node a,node b)
{
	return a.x1<b.x1;
}
int main()
{
	int i,j,k,m,n;
	double t,tt,ww,xx,yy;
	while(scanf("%d",&m)!=EOF)
	{
		if(m==0)
		break;
          for(i=0;i<m;i++)
          {
		     scanf("%lf%lf%lf%lf",&g[i].x1,&g[i].y1,&g[i].x2,&g[i].y2);
		     if(g[i].x1>g[i].x2)
		     {
     			t=g[i].x1;
     			g[i].x1=g[i].x2;
     			g[i].x2=t;
     			t=g[i].y1;
     			g[i].y1=g[i].y2;
     			g[i].y2=t;
     		}
          }
          sort(g,g+m,cmp);
          int sum=0;
          for(i=0;i<m;i++)
          {
          	for(j=i+1;j<m;j++)
          	{
	          	if(g[i].x2<g[j].x1)
	          	break;
	          	/*if(((g[i].y2-g[i].y1)*(g[j].x2-g[j].x1)-(g[i].x2-g[i].x1)*(g[j].y2-g[j].y1))<1e-6)
	               	continue;*/
	          	xx=((g[j].y1*g[j].x2-g[j].y2*g[j].x1)*(g[i].x2-g[i].x1)-(g[i].y1*g[i].x2-g[i].y2*g[i].x1)*(g[j].x2-g[j].x1))/((g[i].y2-g[i].y1)*(g[j].x2-g[j].x1)-(g[j].y2-g[j].y1)*(g[i].x2-g[i].x1));//交点横坐标；
	          	//yy=((g[i].y1*g[i].x2-g[i].y2*g[i].x1)*(g[j].y2-g[j].y1)-(g[j].y1*g[j].x2-g[j].y2*g[j].x1)*(g[i].y2-g[i].y1))/((g[j].x2-g[j].x1)*(g[i].y2-g[i].y1)-(g[i].x2-g[i].x1)*(g[j].y2-g[j].y1));//交点纵坐标，其实可以不求纵坐标的；
	          	if(xx>=g[i].x1&&xx<=g[i].x2&&xx>=g[j].x1&&xx<=g[j].x2)
	          	sum++;
	          }
          }
          printf("%d\n",sum);
	}
	return 0;
}