自创！基于单数组实现prim算法生成MST

#include <iostream>
#include <cmath>
using namespace std;

const int MAXN = 100;
enum Ifexist {R, U}; // Remain-set and U-set
// U集合表示最终为MST节点的集合
// Remain集合表示当前还剩元素
struct Point {
    double x, y;
    Ifexist ife;
};
Point r[MAXN];//REMAIN

bool isempty(int n, Point s[]);// 谓词：判断Remain集合是否为空
inline double calculateD(Point a, Point b);
int main() {
    double x, y;
    int n;
    int keepi;
    double kmin, ans;
    while (cin >> n) {
        ans = 0;
        for (int i = 0; i < n; ++i) {
            cin >> x >> y;
            r[i].x = x;
            r[i].y = y;
            r[i].ife = R;

        }
        r[0].ife = U;
        while (!isempty(n, r)) {
            kmin = 100000;
		  // On^2 计算R和U 集合最近元素距离
            for (int i = 0; i < n; ++i) {
                if (r[i].ife == R) {
                    for (int j = 0; j < n; ++j) {
                        if (r[j].ife == U) {
                            double dst = calculateD(r[i], r[j]);
                            if (dst < kmin) {
                                kmin = dst;
                                keepi = i;
                            }
                        }
                    }
                }
            }
            ans += kmin;// 累加R和U集合最近的边
            r[keepi].ife = U;//R节点数目减一，也意味着U集合加一
        }
        cout.precision(2);
	  //cout<<ans<<endl;
        printf("%.2f\n", ans);
    }
    return 0;


}
bool isempty(int n, Point s[]) {
    for (int i = 0; i < n; ++i) {
        if (s[i].ife == R)
            return false;
    }
    return true;
}

inline double calculateD(Point a, Point b) {
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}