最接近的四点对c程序

Question

我需要找到最接近的四点对 C 程序。这段代码分三点。我需要这个解决方案的四点。

我试过这个。此解决方案适用于三个输入。

当我输入三个点时，我会得到最近的点，但我需要四个点中最接近的点。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct Point
{
    int x, y ;
};
double getDistanceAB(struct Point a, struct Point b)
{
    double distanceAB;
    distanceAB = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) *(a.y-b.y));
    return distanceAB;
}
double getDistanceBC(struct Point b, struct Point c)
{
    double distanceBC;
    distanceBC = sqrt((b.x - c.x) * (b.x - c.x) + (b.y-c.y) *(b.y-c.y));
    return distanceBC;
}
double getDistanceAC(struct Point a, struct Point c)
{
    double distanceAC;
    distanceAC = sqrt((a.x - c.x) * (a.x - c.x) + (a.y-c.y) *(a.y-c.y));
    return distanceAC;
}
int main()
{
    struct Point a, b, c;
    printf("Enter coordinate of points a: ");
    scanf("%d %d", &a.x, &a.y);
    printf("Enter coordinate of points b: ");
    scanf("%d %d", &b.x, &b.y);
    printf("Enter coordinate of points c: ");
    scanf("%d %d", &c.x, &c.y);
    if((getDistanceAB(a,b))>(getDistanceBC(b,c)) && (getDistanceAB(a,b))>(getDistanceBC(a,c)))
    {
        printf("Point A and B are closest.");
    }
    else if((getDistanceBC(b,c))>(getDistanceAC(a,c)) && (getDistanceBC(b,c))>(getDistanceAC(a,b)))
    {
        printf("Point B and C are closest.");
    }
    else if((getDistanceBC(a,c))>(getDistanceAC(a,b)) && (getDistanceBC(a,c))>(getDistanceAC(b,c)))
    {
        printf("Point A and C are closest.");
    }
    else
    {
        printf("All point are same.");
    }
}

Answer 1

首先，更改此:

double getDistanceAB(struct Point a, struct Point b)
{
    double distanceAB;
    distanceAB = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) *(a.y-b.y));
    return distanceAB;
}
double getDistanceBC(struct Point b, struct Point c)
{
    double distanceBC;
    distanceBC = sqrt((b.x - c.x) * (b.x - c.x) + (b.y-c.y) *(b.y-c.y));
    return distanceBC;
}
double getDistanceAC(struct Point a, struct Point c)
{
    double distanceAC;
    distanceAC = sqrt((a.x - c.x) * (a.x - c.x) + (a.y-c.y) *(a.y-c.y));
    return distanceAC;
}

仅此:

   double getDistance(struct Point a, struct Point b)
{
    double distance;
    distance = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) * (a.y-b.y));
    return distance;
}

函数的要点之一是您不必重复代码。

现在您所要做的就是通过为第四个点再添加一次扫描来创建四个点，并将其添加到决策树中。

请记住您所做的决策树的这一点...如果您使用您在原始帖子中使用的相同逻辑检查点“a”是否不是最接近的，则不必比较点“a” ' 再次。