Go: package main

import (
	"bufio"
	"os"
	"strconv"
	"fmt"
)

const (
	MAX = 100
	INF = 1 << 21
	WHITE = 0
	GRAY = 1
	BLACK = 2
)

var (
	n int
	M [MAX][MAX]int
)

func prim() int {
	var u, minv int
	var d [MAX]int
	var p [MAX]int
	var color [MAX]int

	for i := 0; i < n; i++ {
		d[i] = INF
		p[i] = -1
		color[i] = WHITE
	}
	d[0] = 0
	for {
		minv = INF
		u = -1
		for i := 0; i < n; i++ {
			if minv > d[i] && color[i] != BLACK {
				u = i
				minv = d[i]
			}
		}
		if u == -1 {
			break
		}
		color[u] = BLACK
		for v := 0; v < n; v++ {
			if color[v] != BLACK && M[u][v] != INF {
				if d[v] > M[u][v] {
					d[v] = M[u][v]
					p[v] = u
					color[v] = GRAY
				}
			}
		}
	}
	sum := 0
	for i := 0; i < n; i++ {
		if p[i] != -1 {
			sum += M[i][p[i]]
		}
	}
	return sum
}

func main() {
	scanner := bufio.NewScanner(os.Stdin)
	scanner.Split(bufio.ScanWords)
	scanner.Scan()
	n, _ = strconv.Atoi(scanner.Text())
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			scanner.Scan()
			e, _ := strconv.Atoi(scanner.Text())
			if e == -1 {
				M[i][j] = INF
			} else {
				M[i][j] = e
			}
		}
	}
	fmt.Println(prim())
}

C++: #include <iostream>
#define N 100
#define INFINITY 2000000000
using namespace std;
void prim();
int M[N][N],d[N],p[N],u,n,mincost,i,j;
string color[N];
int main(){
  cin >> n;
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      cin >> M[i][j];
    }
  }
  prim();
  return 0;
}
void prim(){
  for(i=0;i<n;i++) color[i]="WHITE",d[i]=INFINITY;
  d[0]=0;
  p[0]=-1;
  while(1){
    mincost=INFINITY;
    for(i=0;i<n;i++){
      if(color[i]!="BLACK"&&d[i]<mincost){
	mincost=d[i];
	u=i;
      }
    }
    if(mincost==INFINITY) break;
    color[u]="BLACK";
    for(i=0;i<n;i++){
      if(color[i]!="BLACK"&&M[u][i]!=-1){
	if(M[u][i]<d[i]){
	  d[i]=M[u][i];
	  p[i]=u;
	  color[i]="GLAY";
	}
      }
    }
  }
  int sum=0;
  for(i=0;i<n;i++) sum+=d[i];
  cout << sum << endl;
}

C++: #include <iostream>
#include <vector>
#include <algorithm>
#define INF 1e9
using namespace std;

int main(){
    //declaration and input
    int n;
    cin >> n;
    vector<vector<int>> graph(n, vector<int>(n));
    for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) cin >> graph[i][j];
    
    //get Minimum Spanning Tree by Prim's algorithm
    vector<int> parent_of(n, -1);
    vector<int> cost(n, INF);
    vector<int> state(n, 0);
    cost[0] = 0;
    state[0]++;

    int sum = 0;
    int parent = -1;
    while (true){
        int min_cost = INF;
        int next = -1;
        for (int i = 0; i < n; i++){
            if (state[i] == 1 && cost[i] < min_cost){
                min_cost = cost[i];
                next = i;
            }
        }
        if (min_cost == INF)
            break;
        sum += cost[next];
        state[next] = 2;
        for (int i = 0; i < n; i++){
            if (state[i] != 2 && graph[i][next] != -1){
                state[i] = 1;
                if (cost[i] > graph[i][next])
                    cost[i] = graph[i][next];
            }
        }
    }

    /*int sum = 0;
    for (int i = 0; i < n; i++){
        if (parent_of[i] != -1)
            sum += graph[i][parent_of[i]];
    }*/

    cout << sum << endl;
    return 0;
}

C++: #include<iostream>
#include<unordered_set>
#define N 105
#define UNDEFINED -1
static const int INF = (1<<21);

int n,a;
int d[N], p[N];
int G[N][N];

int prim() {
    int u, min;
    std::unordered_set<int> hasVisited;
    for(int i = 0; i < N; i++) {
        d[i] = INF;
        p[i] = UNDEFINED;
    }
    d[0] = 0;
    while(true) {
        min = INF;
        u = UNDEFINED;
        for(int i = 0; i < n; i++) {
            if(d[i] < min && hasVisited.find(i) == hasVisited.end()) {
                u = i;
                min = d[i];
            }
        }
        if (u == UNDEFINED) break;
        hasVisited.insert(u);
        for(int i = 0; i < n; i++) {
            if(hasVisited.find(i) == hasVisited.end() && G[u][i] != UNDEFINED) {
                if(d[i] > G[u][i]) {
                    d[i] = G[u][i];
                    p[i] = u;
                }
            }
        }
    }
    int sum = 0;
    for(int i = 0; i < n; i++) {
        if(p[i] != UNDEFINED) {
            sum += G[i][p[i]];
        }
    }
    return sum;
}

int main() {
    std::cin >> n;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {

            std::cin >> a;
            G[i][j] = a;
        }
    }
    std::cout << prim() << std::endl;
    return 0;
}

C++: #include <iostream>
#include <stack>
#define N 100
#define WHITE 0
#define GREY 1
#define BLACK 2
#define INFTY 100000
using namespace std;

int color[N];//color用来记录访问状态
int M[N][N];//M[N][N]是图的邻接矩阵表示法
int d[N],p[N];
//d[N]用来记录连接T内顶点与V-T内顶点的边中，权值最小的边的权值
//p[n]用来记录MST中顶点v的父节点
void init_color_d()
{
    for(int i = 0; i < N; i++)
    {
        color[i] = WHITE;
        d[i] = INFTY;
    }
}
int prim(int n)
{
    init_color_d();
    d[0] = 0;//起始点为0结点，距离设置为0
    p[0] = -1;
    int u;
    while(true)//访问结点为u
    {
        int mincost = INFTY;
        for(int i = 0; i < n; i++)
        {
            if(color[i]!=BLACK&&d[i]<mincost)//求最小值
            {
                mincost = d[i];
                u = i;
            }
        }
        if(mincost==INFTY)break;//这时候表示没有取到最小值，即最小生成树已经完成
        color[u] = BLACK;//因为u是最小值，那么就先标志为访问完
        for(int v = 0; v < n; v++)
        {
            if(color[v]!=BLACK&&M[u][v]!=-1)//如果v没有被选入到MST集合里面，并且此边存在
            {
                if(M[u][v]<d[v])
                {
                    d[v] = M[u][v];//d[v]记录最小值
                    p[v] = u;//记录父节点
                    color[v] = GREY;//设置为已经尝试过此结点
                }
            }
        }
    }
    int sum = 0;
    for(int i = 0;i < n;i++){
        if(p[i]!=-1)sum+=M[i][p[i]];
    }
    return sum;
}

int main()
{
    int n;
    cin >> n;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j <n; j++)
        {
            cin >> M[i][j];
        }
    }
    cout<<prim(n)<<endl;

    return 0;
}

C++: #include <bits/stdc++.h>
using namespace std;
using Graph = vector<vector<int>>;

vector<int> P;

int root(int v){ //根を返す
    if(P[v] == v) return v;
    else return root(P[v]);
}

bool is_same_set(int a, int b){ // 同じならtrue
    return root(a) == root(b);
}

void unite(int a, int b){ // aの根をbの根にくっつける
    P[root(a)] = root(b);
}

int main(){
    int n; cin >> n;
    P.resize(n); //　Pは作りかけの森を想定する
    for(int i = 0; i < n; i++){
        P[i] = i;
    }
    vector<tuple<int, int, int>> vec; //<コスト,始点, 終点>を格納する配列
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            int a; cin >> a;
            if(a >= 0 && i < j){
                vec.push_back(make_tuple(a, i, j));
            }
        }
    }
    sort(vec.begin(), vec.end()); //コスト順にソート
    int ans = 0;
    for(auto v: vec){
        int cost = get<0>(v), str = get<1>(v), end = get<2>(v);
        if(is_same_set(str, end)) continue; //繋がっている
        //もし繋がっていなければ
        ans += cost; // コストを足す
        unite(str, end); //つなげてあげる
    }
    cout << ans << endl;
}