hdu1086

You can Solve a Geometry Problem too

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 4

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

20.00 0.00 1.00 1.000.00 1.00 1.00 0.0030.00 0.00 1.00 1.000.00 1.00 1.00 0.0000.00 0.00 1.00 0.000

 

Sample Output

13
这是一道简单的几何求线段相交的问题，似乎在这道题目里边没有出现多个顶点交于一点的情况，所以处理起来会相对的简单一些的，
需要注意的是，在判断线段相交时，运用叉积操作时要判断两侧相交才能保证，线段相交

 代码如下，

#include <iostream>#include <cstdio>using namespace std;const int maxn=102;struct node{    double x1,y1;    double x2,y2;};struct dot{    int x,y;};node a[maxn];int cross(node a,node b){   if(((a.x2-a.x1)*(b.y1-a.y1)-(a.y2-a.y1)*(b.x1-a.x1))*((a.x2-a.x1)*(b.y2-a.y1)-(a.y2-a.y1)*(b.x2-a.x1))>0)    return 0;   if(((b.x2-b.x1)*(a.y1-b.y1)-(b.y2-b.y1)*(a.x1-b.x1))*((b.x2-b.x1)*(a.y2-b.y1)-(b.y2-b.y1)*(a.x2-b.x1))>0)    return 0;   return 1;}int main(){    int n;    while(scanf("%d",&n)!=EOF&&n)    {        int sum=0;         for(int i=0;i<n;i++)        {            scanf("%lf%lf%lf%lf",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2);        }        for(int j=0;j<n-1;j++)            for(int i=j+1;i<n;i++)            {            if(cross(a[i],a[j]))                sum++;            }            printf("%d\n",sum);    }    return 0;}