(hdu step 7.1.2)You can Solve a Geometry Problem too(乞讨n条线段,相交两者之间的段数)

称号：

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：推断两条线断是否相交。

代码例如以下：

#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;
}

版权声明：本文博主原创文章，博客，未经同意不得转载。