hdoj 1086 You can Solve a Geometry Problem too （判断线段相交）

You can Solve a Geometry Problem too

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5298    Accepted Submission(s): 2526

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1086

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 0.00

0.00 0.00 1.00 0.00

0

 

Sample Output

1

3

 

题意 ：给你 n 条线段，求交点个数。

解题思路：http://www.cnblogs.com/Duahanlang/archive/2013/05/11/3073434.html

AC 代码：

#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>

using namespace std;

const int maxv = 105;
struct point {          // 点
    double x;           // x 坐标
    double y;           // y 坐标
} first[maxv], final[maxv];

double intersect(point p1, point q1, point p2)
{
    point tmp1, tmp2;
    tmp1.x = q1.x - p1.x, tmp1.y = q1.y - p1.y;     // tmp1 = q1-p1   是向量
    tmp2.x = p2.x - p1.x, tmp2.y = p2.y - p1.y;     // tmp2 = p2-p1   是向量
    double ans = tmp1.x * tmp2.y - tmp2.x * tmp1.y; // ans = (q1-p1) x (p2-p1) (向量叉积)
    return ans;
}

int main()
{
    int n, i, j;
    double ans1, ans2, ans3, ans4;
    int cnt;                //相交次数
    while(cin >> n && n) {
        cnt = 0;
        memset(first, 0, sizeof(first));
        memset(final, 0, sizeof(final));
        for(i = 0; i < n; ++i) {
            cin >> first[i].x >> first[i].y >> final[i].x >> final[i].y;
        }
        for(i = 0; i < n; ++i) {
            for(j = i+1; j < n; ++j) {
                ans1 = intersect(first[i], final[i], first[j]);
                ans2 = intersect(first[i], final[i], final[j]);
                double t1 = ans1 * ans2;
                ans3 = intersect(first[j], final[j], first[i]);
                ans4 = intersect(first[j], final[j], final[i]);
                double t2 = ans3 * ans4;
                if(t1 <= 0 && t2 <= 0)
                    cnt++;
            }
        }
        cout << cnt << endl;
    }
    return 0;
}