hdu 1086 You can Solve a Geometry Problem too(求线段相交点个数 模板)
2014-07-26 20:55
447 查看
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1086
Total Submission(s): 7227 Accepted Submission(s): 3504
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
Sample Output
Author
lcy
套用吉大模板;
代码如下:
You can Solve a Geometry Problem too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7227 Accepted Submission(s): 3504
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
套用吉大模板;
代码如下:
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; const double eps=1e-10; struct point { double x, y; }; double min(double a, double b) { return a < b ? a : b; } double max(double a, double b) { return a > b ? a : b; } bool inter(point a, point b, point c, point d) { if (min(a.x, b.x) > max(c.x, d.x)|| min(a.y, b.y) > max(c.y, d.y)|| min(c.x, d.x) > max(a.x, b.x)||min(c.y, d.y) > max(a.y, b.y) ) return 0; double h, i, j, k; h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x); i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x); j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x); k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x); return h * i <= eps && j * k <= eps; } int main() { int n; int i, j; point p[117][2]; while(cin>>n && n) { for(i = 0; i < n; i++) { cin>>p[i][0].x>>p[i][0].y>>p[i][1].x>>p[i][1].y; } int count = 0; for(i = 0; i < n; i++) { for(j = i+1; j < n; j++) { if(inter(p[i][0],p[i][1],p[j][0],p[j][1])) count++; } } cout<<count<<endl; } return 0; }
相关文章推荐
- 【线段相交】(吉大模板):hdu 1086 You can Solve a Geometry Problem too
- hdu 1086:You can Solve a Geometry Problem too(计算几何,判断两线段相交,水题)
- hdu 1086 You can Solve a Geometry Problem too(线段相交的交点个数)
- HDU 1086 You can Solve a Geometry Problem too(简单的线段相交)
- HDU 1086 You can Solve a Geometry Problem too 判断任意两线段是否相交
- HDU 1086 You can Solve a Geometry Problem too(判定线段相交 规范相交和非规范相交)
- HDU 1086 You can Solve a Geometry Problem too (判断线段相交)
- hdu-1086 You can Solve a Geometry Problem too(线段是否相交)
- HDU 1086 You can Solve a Geometry Problem too(判定线段相交)
- HDU-#1086 You can Solve a Geometry Problem too(线段相交判定)
- HDU 1086.You can Solve a Geometry Problem too【判断两线段相交】【数学题】【12月30】
- hdu 1086 You can Solve a Geometry Problem too(线段相交+枚举)
- Hdu 1086 You can Solve a Geometry Problem too[判断线段相交,完整版]
- hdu 1086 You can Solve a Geometry Problem too(线段相交点的个数)
- hdu 1086 You can Solve a Geometry Problem too 线段相交
- hdu_1086 You can Solve a Geometry Problem too(线段相交)
- 杭电hdu 1086 You can Solve a Geometry Problem too 线段相交
- HDU 1086 You can Solve a Geometry Problem too(判断两线段是否相交)跨立实验
- HDU 1086 You can Solve a Geometry Problem too(判断线段是否相交,非规范相交)
- HDU 1086 You can Solve a Geometry Problem too(水题,判断线段相交)