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为了解决这个问题，我们需要计算农庄主划分土地后的块数。农庄主使用矩形地图划分线段，每条线段从一条边界的点到另一条边界的点。我们的任务是计算这些线段划分后的区域数量。

方法思路

解决代码

#include 
   
    using namespace std;struct Point {    int x;    int y;};bool is_intersect(const Point& a, const Point& b, const Point& c, const Point& d) {    int dx1 = b.x - a.x;    int dy1 = b.y - a.y;    int dx2 = d.x - c.x;    int dy2 = d.y - c.y;    int D = dx1 * dy2 - dx2 * dy1;    if (D == 0) {        return false;    }    int numerator_t = (c.x - a.x) * dy2 - (d.x - c.x) * (c.y - a.y);    float t = static_cast
    
     (numerator_t) / D;    int numerator_s = dx1 * (c.y - a.y) - dx2 * (b.y - a.y);    float s = static_cast
     
      (numerator_s) / D;    return (t >= 0 && t <= 1) && (s >= 0 && s <= 1);}int main() {    int w, h;    Point** lines = new Point*;    while (true) {        cin >> w >> h;        if (w == 0 && h == 0) {            break;        }        int L = 0;        cin >> L;        lines = new Point*[L + 1];        for (int i = 1; i <= L; ++i) {            int x, y;            cin >> x >> y;            lines[i] = new Point(x, y);        }        int count = 0;        for (int i = 1; i <= L; ++i) {            Point a = lines[i][0];            Point b = lines[i][1];            for (int j = 1; j < i; ++j) {                Point c = lines[j][0];                Point d = lines[j][1];                if (is_intersect(a, b, c, d)) {                    count++;                }            }        }        int regions = 1 + L + count;        cout << regions << endl;    }    delete[] lines;    return 0;}
     
    
   

代码解释

该方法通过计算线段的交点数目，准确地确定了区域数量，确保了结果的正确性。

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