(hdu step 7.1.2)You can Solve a Geometry Problem too(求n条线段中,线段两两相交的数量)_计算机图形学编写一个类或结构体，表示平面上的一条线段，并据此计算并输出线段的长-优快云博客

本文链接：https://blog.youkuaiyun.com/hjd_love_zzt/article/details/44339467

本文介绍了一个简单的几何问题：计算给定线段集合中所有交点的数量，并提供了完整的C++实现代码。该问题要求考虑线段可能在同一点相交的情况。

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题目：

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 145 Accepted Submission(s): 100

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

Author

lcy

题目分析：

简单题。核心还是：判断两条线断是否相交。

代码如下：

#include<iostream>
#include <cstdio>

using namespace std;
struct Line{
    double x1,y1,x2,y2;
}lines[110];

bool isCross(Line a,Line b){
    if(((a.x1-b.x1)*(a.y2-b.y1)-(a.x2-b.x1)*(a.y1-b.y1))*((a.x1-b.x2)*(a.y2-b.y2)-(a.x2-b.x2)*(a.y1-b.y2))>0)return false;
    if(((b.x1-a.x1)*(b.y2-a.y1)-(b.x2-a.x1)*(b.y1-a.y1))*((b.x1-a.x2)*(b.y2-a.y2)-(b.x2-a.x2)*(b.y1-a.y2))>0)return false;
    return true;
}


int main(){
	int n;
	while(scanf("%d",&n)!=EOF,n){
		int i;
		for(i = 0 ; i < n ; ++i){
			scanf("%lf %lf %lf %lf",&lines[i].x1,&lines[i].y1,&lines[i].x2,&lines[i].y2);
		}

		int j;
		int ans = 0;
		for(i = 0 ; i < n ; ++i){//求n条线段中,相交线段的数量
			for(j = i+1; j < n ; ++j){
				if(isCross(lines[i],lines[j]) == true){
					ans++;
				}
			}
		}

		printf("%d\n",ans);
	}

	return 0;
}