(叉积,线段判交)HDU1086 You can Solve a Geometry Problem too
You can Solve a Geometry Problem too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12959 Accepted Submission(s): 6373
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.
A test case starting with 0 terminates the input and this test case is not to be processed.
叉积求线段判交的参考链接:
https://www.cnblogs.com/Duahanlang/archive/2013/05/11/3073434.html
https://www.cnblogs.com/tuyang1129/p/9390376.html
C++代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct Point{
double x1,y1,x2,y2;
}node[];
int cross(const Point &a, const Point &b){
double k1 = (a.x2 - a.x1) * (b.y1 - a.y1) - (a.y2 - a.y1) * (b.x1 - a.x1);
double k2 = (a.x2 - a.x1) * (b.y2 - a.y1) - (a.y2 - a.y1) * (b.x2 - a.x1);
if(k1 * k2 <= ){
return ;
}
else
return ;
}
int main(){
int n;
while(scanf("%d",&n),n){
int ans = ;
for(int i = ; i < n; i++){
scanf("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2);
}
for(int i = ; i < n-; i++){
for(int j = i + ; j < n; j++){
ans += (cross(node[i],node[j])) && (cross(node[j],node[i]));
}
}
printf("%d\n",ans);
}
return ;
}
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