给定5个线段的端点坐标，能够组成几个封闭图形，用C++代码实现

下面是一个用于计算给定5个线段能够组成多少个封闭图形的C++代码示例：

#include <iostream>
#include <vector>
#include <algorithm>

struct Point {
    double x, y;
};

struct LineSegment {
    Point start, end;
};

bool areSegmentsIntersecting(LineSegment segment1, LineSegment segment2) {
    double x1 = segment1.start.x;
    double y1 = segment1.start.y;
    double x2 = segment1.end.x;
    double y2 = segment1.end.y;

    double x3 = segment2.start.x;
    double y3 = segment2.start.y;
    double x4 = segment2.end.x;
    double y4 = segment2.end.y;

    double dx1 = x2 - x1;
    double dy1 = y2 - y1;
    double dx2 = x4 - x3;
    double dy2 = y4 - y3;

    double denominator = dx1 * dy2 - dx2 * dy1;

    if (denominator == 0) {
        return false;  // 线段平行或共线
    }

    double t1 = ((x3 - x1) * dy2 + (x1 - x2) * dy1) / denominator;
    double t2 = ((y1 - y3) * dx2 + (x3 - x4) * dy2) / denominator;

    if (t1 >= 0 && t1 <= 1 && t2 >= 0 && t2 <= 1) {
        return true;  // 线段相交
    }

    return false;  // 线段不相交
}

int countClosedShapes(std::vector<LineSegment> segments) {
    int count = 0;
    int n = segments.size();

    for (int i = 0; i < n; i++) {
        bool isClosed = true;

        for (int j = 0; j < n; j++) {
            if (i == j) {
                continue;
            }

            if (areSegmentsIntersecting(segments[i], segments[j])) {
                isClosed = false;
                break;
            }
        }

        if (isClosed) {
            count++;
        }
    }

    return count;
}

int main() {
    std::vector<LineSegment> segments;

    // 假设已知的5个线段的端点坐标为:
    Point a = {0, 0};
    Point b = {1, 1};
    Point c = {1, 1};
    Point d = {2, 2};
    Point e = {2, 2};
    Point f = {3, 3};
    Point g = {3, 3};
    Point h = {4, 4};

    LineSegment segment1 = {a, b};
    LineSegment segment2 = {c, d};
    LineSegment segment3 = {e, f};
    LineSegment segment4 = {g, h};
    LineSegment segment5 = {a, h};

    segments.push_back(segment1);
    segments.push_back(segment2);
    segments.push_back(segment3);
    segments.push_back(segment4);
    segments.push_back(segment5);

    int closedShapesCount = countClosedShapes(segments);

    std::cout << "Number of closed shapes: " << closedShapesCount << std::endl;

    return 0;
}

在上述代码中，我们首先定义了 `Point` 和 `LineSegment` 的结构体，分别表示点和线段。然后，我们实现了 `areSegmentsIntersecting` 函数来判断两条线段是否相交。接下来，我们使用 `countClosedShapes` 函数来计算给定线段能够组成的封闭图形数量。最后，在 `main` 函数中，我们定义了5个线段的端点坐标，并将其传递给 `countClosedShapes` 函数来得到结果。

运行代码后，会输出给定线段能够组成的封闭图形的数量。