51Nod－1264－线段相交

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ACM模版

描述

给出平面上两条线段的两个端点，判断这两条线段是否相交（有一个公共点或有部分重合认为相交）。如果相交，输出”Yes”，否则输出”No”。

Input
第1行：一个数T，表示输入的测试数量(1 <= T <= 1000)
第2 - T + 1行：每行8个数，x1,y1,x2,y2,x3,y3,x4,y4。(-10^8 <= xi, yi <= 10^8)
(直线1的两个端点为x1,y1 | x2, y2,直线2的两个端点为x3,y3 | x4, y4)

Output
输出共T行，如果相交输出”Yes”，否则输出”No”。

Input示例
2
1 2 2 1 0 0 2 2
-1 1 1 1 0 0 1 -1

Output示例
Yes
No

题解

计算几何相关问题。

代码

#include <iostream>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 argc, const char * argv[]){#ifndef TEST_DATA    freopen("input.txt", "r", stdin);//    freopen("output.txt", "w", stdout);#endif    int T;    cin >> T;    point a, b, c, d;    while (T--)    {        cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y >> d.x >> d.y;        if (inter(a, b, c, d))        {            cout << "Yes\n";        }        else        {            cout << "No\n";        }    }    return 0;}

参考

《判断线段相交》

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