You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2734 Accepted Submission(s): 1284

Problem Description

Many geometry（幾何）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（線段）, please output the number of all intersections（交點(diǎn)）. 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 詳細(xì)看 《計(jì)算幾何—基礎(chǔ)》 1 #include<iostream>
2 using namespace std;
3 double p[100][2],q[100][2];
4 double direction( double p[] , double q[] , double r[])
5 {
6 return ( (r[0] - p[0]) * (q[1] - p[1]) - (r[1] - p[1]) * (q[0] - p[0]));
7 }
8 bool onsegment(double p[],double q[], double r[])
9 {
10 if(((r[0] - p[0])*(r[0] - q[0]) <= 0) && ((r[1] - p[1]) * (r[1] - q[1]) <= 0))
11 return true;
12 else return false;
13 }
14 bool judge(int i, int j)
15 {
16 double d1 , d2 , d3 , d4 ;
17 d1 = direction(p[i],q[i],p[j]);
18 d2 = direction(p[i],q[i],q[j]);
19 d3 = direction(p[j],q[j],p[i]);
20 d4 = direction(p[j],q[j],q[i]);
21 if( (d1 * d2 < 0) && (d3 * d4 < 0))
22 return true;
23 else if( d1 == 0 && onsegment(p[i],q[i],p[j]) == 1)
24 return true;
25 else if( d2 == 0 && onsegment(p[i],q[i],q[j]) == 1)
26 return true;
27 else if( d3 == 0 && onsegment(p[j],q[j],p[i]) == 1)
28 return true;
29 else if( d4 == 0 && onsegment(p[j],q[j],q[i]) == 1)
30 return true;
31 else return false;
32 }
33 int main(){
34 int n, i , j , count;
35 while( scanf( "%d" , &n ) , n)
36 {
37 for( i = 0 ; i < n ; ++ i )
38 {
39 scanf( "%lf %lf %lf %lf", &p[i][0] , &p[i][1] , &q[i][0] , &q[i][1] ) ;
40 }
41 for( i = 0 , count = 0 ; i < n ; i ++ )
42 for( j = i + 1 ; j < n ; j ++ )
43 if( judge( i , j ) == 1 )
44 ++count;
45 printf("%d\n",count);
46 }
47 return 0;
48 }

　　

總結(jié)

以上是生活随笔為你收集整理的HDU 1086 You can Solve a Geometry Problem too的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。