用分治法解决凸包问题有可视化c语言

以下是使用C语言实现分治法解决凸包问题的示例代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <graphics.h>

typedef struct point {
    int x;
    int y;
} Point;

typedef struct line {
    Point p1;
    Point p2;
} Line;

void drawLine(Line l) {
    line(l.p1.x, l.p1.y, l.p2.x, l.p2.y);
}

float distance(Point p1, Point p2) {
    return sqrt(pow(p1.x - p2.x, 2) + pow(p1.y - p2.y, 2));
}

float crossProduct(Point p1, Point p2, Point p3) {
    return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y);
}

void drawConvexHull(Point *points, int n) {
    if (n <= 1) return;

    Point leftmost = points[0];
    Point rightmost = points[0];
    for (int i = 1; i < n; i++) {
        if (points[i].x < leftmost.x) {
            leftmost = points[i];
        }
        if (points[i].x > rightmost.x) {
            rightmost = points[i];
        }
    }

    setcolor(RED);
    drawLine((Line) {leftmost, rightmost});

    Point *leftPoints = (Point *) malloc((n + 1) * sizeof(Point));
    Point *rightPoints = (Point *) malloc((n + 1) * sizeof(Point));
    int leftCount = 0;
    int rightCount = 0;

    for (int i = 0; i < n; i++) {
        if (crossProduct(leftmost, rightmost, points[i]) > 0) {
            leftPoints[leftCount++] = points[i];
        } else if (crossProduct(leftmost, rightmost, points[i]) < 0) {
            rightPoints[rightCount++] = points[i];
        }
    }

    drawConvexHull(leftPoints, leftCount);

    drawConvexHull(rightPoints, rightCount);

    free(leftPoints);
    free(rightPoints);
}

int main() {
    int gd = DETECT, gm;
    initgraph(&gd, &gm, "");

    Point points[] = {{50, 50}, {100, 100}, {150, 50}, {200, 200}, {250, 50}, {300, 100}, {350, 50}, {400, 150}};
    int n = sizeof(points) / sizeof(points[0]);

    drawConvexHull(points, n);

    getch();
    closegraph();
    return 0;
}

这段代码使用了图形库来进行可视化，可以在运行程序时看到凸包的绘制过程。具体实现思路是使用分治法将点集分为左右两部分，然后递归地求解左右两部分的凸包，最后将左右两个凸包合并成一个凸包。在合并时，找到左边凸包的最右边的点和右边凸包的最左边的点，然后连接这两个点，将它们之间的线段作为凸包的边。