HDU1086You can Solve a Geometry Problem too （斜率问题）

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8605    Accepted Submission(s): 4194

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

 

思路：排除两线段没有相交可能的情况，即线段的最大点比另一条线段的最小点还小

然后两线段相交要求   线段1斜率介于线段1的一端点与另一条线段2的俩端点的斜率之间

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
struct node
{
    double x;
    double y;
};
double chaji(node a,node b,node c)    //两条线段斜率的差，double !!!
{
    return (b.y-a.y)*(c.x-a.x)-(c.y-a.y)*(b.x-a.x);
}
int count(node a1,node a2,node b1,node b2)
{
    if(max(a1.x,a2.x)<min(b1.x,b2.x))
        return 0;
    if(max(a1.y,a2.y)<min(b1.y,b2.y))
      #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
struct node
{
    double x;
    double y;
};
double chaji(node a,node b,node c)    //两条线段斜率的差，double !!!
{
    return (b.y-a.y)*(c.x-a.x)-(c.y-a.y)*(b.x-a.x);
}
int count(node a1,node a2,node b1,node b2)
{
    if(max(a1.x,a2.x)<min(b1.x,b2.x))
        return 0;
    if(max(a1.y,a2.y)<min(b1.y,b2.y))
        return 0;
    if(min(a1.x,a2.x)>max(b1.x,b2.x))
        return 0;
    if(min(a1.y,a2.y)>max(b1.y,b2.y))
        return 0;
 
      if(chaji(a1,a2,b1)*chaji(a1,b2,a2)<0)
      return 0;
      if(chaji(b1,b2,a1)*chaji(b1,a2,b2)<0)
      return 0;
    return 1;
}
int main()
{
    int n,t;
    node nodea[105],nodeb[105];
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        t=0;
        for(int i=1; i<=n; i++)
        {
            scanf("%lf%lf%lf%lf",&nodea[i].x,&nodea[i].y,&nodeb[i].x,&nodeb[i].y);
        }
        for(int i=1; i<n; i++)
        {
            for(int j=i+1; j<=n; j++)
            {
               // printf("!!!!\n");
                if(count(nodea[i],nodeb[i],nodea[j],nodeb[j]))
                {
                  //  printf("@222\n");
                    t++;
                }
 
            }
        }
        printf("%d\n",t);
    }
 
    return 0;
}