题目：
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 |
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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 :) Note: |
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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. |
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Output For each case, print the number of intersections, and one line one case. |
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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 |
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Sample Output 1 3 |
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Author lcy |
题目分析：
简单题。核心还是：判断两条线断是否相交。
代码如下：
#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;
}