hdoj1086You can Solve a Geometry Problem too【线段相交判断】_综合

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8964 Accepted Submission(s): 4375

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

题意：给定线段两端坐标计算这些线段的交点个数可以重复计算

解题思路：枚举每两条线段计算交点

#include<cstdio>
#include<cstdlib>
#include<cstring>
#define eps 1e-10
using namespace std;
double MAX(double a,double b){return a>b?a:b;
}
double MIN(double a,double b){return a<b?a:b;
}
struct point{double x,y;
};
struct line{point a,b;
}A[110];
bool judge(int a,int b){      if(MIN(A[a].a.x,A[a].b.x)>MAX(A[b].a.x,A[b].b.x)||MIN(A[a].a.y,A[a].b.y)>MAX(A[b].a.y,A[b].b.y)||MIN(A[b].a.x,A[b].b.x)>MAX(A[a].a.x,A[a].b.x)||MIN(A[b].a.y,A[b].b.y)>MAX(A[a].a.y,A[a].b.y))      return false;      double h,i,j,k;      h=(A[a].b.x-A[a].a.x)*(A[b].a.y-A[a].a.y)-(A[a].b.y-A[a].a.y)*(A[b].a.x-A[a].a.x);      i=(A[a].b.x-A[a].a.x)*(A[b].b.y-A[a].a.y)-(A[a].b.y-A[a].a.y)*(A[b].b.x-A[a].a.x);      j=(A[b].b.x-A[b].a.x)*(A[a].a.y-A[b].a.y)-(A[b].b.y-A[b].a.y)*(A[a].a.x-A[b].a.x);      k=(A[b].b.x-A[b].a.x)*(A[a].b.y-A[b].a.y)-(A[b].b.y-A[b].a.y)*(A[a].b.x-A[b].a.x);      return h*i<=eps&&j*k<=eps;      
}  
int main()
{int n,i,j,k,ans;while(scanf("%d",&n),n){for(i=0;i<n;++i){scanf("%lf%lf%lf%lf",&A[i].a.x,&A[i].a.y,&A[i].b.x,&A[i].b.y);}ans=0;for(i=0;i<n;++i){for(j=i+1;j<n;++j){if(judge(i,j))ans++;}}printf("%d\n",ans);}return 0;
}