用c++的remove_if写一个数组中所有线段元素中存在相交的线段的元素被移除，但要获取他们的交点存到一个新的数组

好的，我来为您解答这个问题。

首先，我们需要定义一个结构体来表示线段，包含起点和终点坐标：

struct Line {
    int x1, y1, x2, y2; // 起点和终点坐标
};

接着，我们需要判断两条线段是否相交，可以使用叉积来判断：

// 判断两条线段是否相交
bool isIntersect(Line l1, Line l2) {
    int c1 = cross(l2.x1-l1.x1, l2.y1-l1.y1, l1.x2-l1.x1, l1.y2-l1.y1);
    int c2 = cross(l2.x2-l1.x1, l2.y2-l1.y1, l1.x2-l1.x1, l1.y2-l1.y1);
    int c3 = cross(l1.x1-l2.x1, l1.y1-l2.y1, l2.x2-l2.x1, l2.y2-l2.y1);
    int c4 = cross(l1.x2-l2.x1, l1.y2-l2.y1, l2.x2-l2.x1, l2.y2-l2.y1);
    return (c1 * c2 < 0) && (c3 * c4 < 0);
}

// 计算叉积
int cross(int x1, int y1, int x2, int y2) {
    return x1 * y2 - x2 * y1;
}

接下来，我们可以使用 `remove_if` 函数来移除所有相交的线段，并将它们的交点存储到一个新的数组中：

// 定义存储交点的结构体
struct Point {
    int x, y;
};

// 移除相交线段，获取交点
void removeIntersect(Line* lines, int n, Line* result, int& m, Point* points, int& k) {
    // 遍历所有线段，判断是否相交
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isIntersect(lines[i], lines[j])) { // 如果相交
                // 计算交点
                Point p = {
                    cross(cross(lines[i].x1, lines[i].y1, lines[i].x2, lines[i].y2),
                          lines[i].x1 - lines[i].x2,
                          cross(lines[j].x1, lines[j].y1, lines[j].x2, lines[j].y2),
                          lines[j].x1 - lines[j].x2)
                          / cross(lines[i].x1 - lines[i].x2, lines[i].y1 - lines[i].y2,
                                   lines[j].x1 - lines[j].x2, lines[j].y1 - lines[j].y2),
                    cross(cross(lines[i].x1, lines[i].y1, lines[i].x2, lines[i].y2),
                          lines[i].y1 - lines[i].y2,
                          cross(lines[j].x1, lines[j].y1, lines[j].x2, lines[j].y2),
                          lines[j].y1 - lines[j].y2)
                          / cross(lines[i].x1 - lines[i].x2, lines[i].y1 - lines[i].y2,
                                   lines[j].x1 - lines[j].x2, lines[j].y1 - lines[j].y2)
                };
                // 将交点存储到数组中
                points[k++] = p;
                // 将相交的线段从原数组中移除
                lines[i] = {0, 0, 0, 0};
                lines[j] = {0, 0, 0, 0};
            }
        }
    }
    // 将不相交的线段存储到结果数组中
    m = 0;
    for (int i = 0; i < n; i++) {
        if (lines[i].x1 != 0 || lines[i].y1 != 0 || lines[i].x2 != 0 || lines[i].y2 != 0) {
            result[m++] = lines[i];
        }
    }
}

完整代码如下：

#include <iostream>
#include <algorithm>

using namespace std;

// 定义存储线段的结构体
struct Line {
    int x1, y1, x2, y2; // 起点和终点坐标
};

// 定义存储交点的结构体
struct Point {
    int x, y;
};

// 判断两条线段是否相交
bool isIntersect(Line l1, Line l2) {
    int c1 = cross(l2.x1-l1.x1, l2.y1-l1.y1, l1.x2-l1.x1, l1.y2-l1.y1);
    int c2 = cross(l2.x2-l1.x1, l2.y2-l1.y1, l1.x2-l1.x1, l1.y2-l1.y1);
    int c3 = cross(l1.x1-l2.x1, l1.y1-l2.y1, l2.x2-l2.x1, l2.y2-l2.y1);
    int c4 = cross(l1.x2-l2.x1, l1.y2-l2.y1, l2.x2-l2.x1, l2.y2-l2.y1);
    return (c1 * c2 < 0) && (c3 * c4 < 0);
}

// 计算叉积
int cross(int x1, int y1, int x2, int y2) {
    return x1 * y2 - x2 * y1;
}

// 移除相交线段，获取交点
void removeIntersect(Line* lines, int n, Line* result, int& m, Point* points, int& k) {
    // 遍历所有线段，判断是否相交
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isIntersect(lines[i], lines[j])) { // 如果相交
                // 计算交点
                Point p = {
                    cross(cross(lines[i].x1, lines[i].y1, lines[i].x2, lines[i].y2),
                          lines[i].x1 - lines[i].x2,
                          cross(lines[j].x1, lines[j].y1, lines[j].x2, lines[j].y2),
                          lines[j].x1 - lines[j].x2)
                          / cross(lines[i].x1 - lines[i].x2, lines[i].y1 - lines[i].y2,
                                   lines[j].x1 - lines[j].x2, lines[j].y1 - lines[j].y2),
                    cross(cross(lines[i].x1, lines[i].y1, lines[i].x2, lines[i].y2),
                          lines[i].y1 - lines[i].y2,
                          cross(lines[j].x1, lines[j].y1, lines[j].x2, lines[j].y2),
                          lines[j].y1 - lines[j].y2)
                          / cross(lines[i].x1 - lines[i].x2, lines[i].y1 - lines[i].y2,
                                   lines[j].x1 - lines[j].x2, lines[j].y1 - lines[j].y2)
                };
                // 将交点存储到数组中
                points[k++] = p;
                // 将相交的线段从原数组中移除
                lines[i] = {0, 0, 0, 0};
                lines[j] = {0, 0, 0, 0};
            }
        }
    }
    // 将不相交的线段存储到结果数组中
    m = 0;
    for (int i = 0; i < n; i++) {
        if (lines[i].x1 != 0 || lines[i].y1 != 0 || lines[i].x2 != 0 || lines[i].y2 != 0) {
            result[m++] = lines[i];
        }
    }
}

int main() {
    // 定义测试用例
    Line lines[] = {
        {0, 0, 1, 1},
        {1, 0, 0, 1},
        {0, 0, 0, 1},
        {0, 1, 1, 1},
        {1, 1, 1, 0},
        {1, 0, 1, 1},
        {0, 0, 1, 0},
        {0, 0, 1, 1}
    };
    int n = sizeof(lines) / sizeof(lines[0]);
    Line result[n];
    int m;
    Point points[n];
    int k = 0;

    // 移除相交线段，获取交点
    removeIntersect(lines, n, result, m, points, k);

    // 输出结果
    cout << "相交线段已移除，不相交线段如下：" << endl;
    for (int i = 0; i < m; i++) {
        cout << "(" << result[i].x1 << "," << result[i].y1 << ") -> (" << result[i].x2 << "," << result[i].y2 << ")" << endl;
    }
    cout << "交点如下：" << endl;
    for (int i = 0; i < k; i++) {
        cout << "(" << points[i].x << "," << points[i].y << ")" << endl;
    }

    return 0;
}