Go: package main

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

//util
var sc = bufio.NewScanner(os.Stdin)

func nextStr() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func printList(list []int) {
	n := len(list)
	buf := bufio.NewWriter(os.Stdout)
	for _, v := range list[0 : n-1] {
		buf.Write([]byte(fmt.Sprintf("%d ", v)))
	}
	buf.Write([]byte(fmt.Sprintf("%d\n", list[n-1])))
	buf.Flush()
}

func max(a, b int) int {
	if a > b {
		return a
	} else {
		return b
	}
}

func min(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}

//solve
const (
	MAX   = 100
	INFTY = (1 << 21)
	WHITE = 0
	GRAY  = 1
	BLACK = 2
)

//V: グラフの頂点全体の集合
//T: MSTに属する頂点の集合
var (
	n     int
	M     [MAX][MAX]int //重みを記録する隣接行列
	color [MAX]int
	p     [MAX]int //MSTにおける頂点の親を記録
	d     [MAX]int //Tに属する頂点と、V−Tに属する頂点を繋ぐ辺の中で、重みが最小のものを記録
)

func prim() int {
	var u, minv int
	//初期化
	for i := 0; i < n; i++ {
		d[i] = INFTY
		p[i] = -1
		color[i] = WHITE
	}

	d[0] = 0

	for {
		//頂点uを決めていく
		//つまり、Tに属する頂点と、V−Tに属する頂点を繋ぐ辺の中で、重みが最小のものを決める
		minv = INFTY
		u = -1
		for i := 0; i < n; i++ {
			//未訪問かつ、辺の重みが最小となるuを選ぶ
			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++ {
			//未訪問かつ、辺が存在するとき
			if color[v] != BLACK && M[u][v] != INFTY {
				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() {
	sc.Split(bufio.ScanWords)
	n = nextInt()

	for i := 0; i < n; i++ {

		for j := 0; j < n; j++ {
			tmp := nextInt()
			if tmp == -1 {
				tmp = INFTY
			}
			M[i][j] = tmp
		}

	}

	fmt.Println(prim())

}

C++: //Kruskal's algorithm
#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

struct E
{
    int weight;
    int begin;
    int end;
};

struct Unionfind
{
    vector<int> parents;

    Unionfind(int n) : parents(n)
    {
        for (int i = 0; i < n; i++)
        {
            parents[i] = i;
        }
    }

    int root(int x)
    {
        if (parents[x] == x)
        {
            return x;
        }
        return parents[x] = root(parents[x]);
    }

    void unite(int x, int y)
    {
        int par_x = root(x);
        int par_y = root(y);
        if (par_x == par_y)
        {
            return;
        }
        parents[par_x] = par_y;
    }

    bool same(int x, int y)
    {
        int par_x = root(x);
        int par_y = root(y);
        return par_x == par_y;
    }
};

int main()
{
    int n;
    cin >> n;
    struct Unionfind tree(n);

    vector<E> e(n * (n - 1) / 2);
    int index = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            int temp;
            cin >> temp;
            if (i < j)
            {
                e[index].weight = temp;
                e[index].begin = i;
                e[index].end = j;
                index++;
            }
        }
    }

    sort(e.begin(), e.end(),
         [](const E &l, const E &r) {
             return l.weight < r.weight;
         });

    int ans = 0;
    for (int i = 0; i < e.size(); i++)
    {
        if (e[i].weight == -1)
        {
            continue;
        }

        if (tree.same(e[i].begin, e[i].end))
        {
            continue;
        }
        else
        {
            tree.unite(e[i].begin, e[i].end);
            ans += e[i].weight;
        }
    }

    cout << ans << endl;
}

/*

<INPUT>
5
 -1 2 3 1 -1
 2 -1 -1 4 -1
 3 -1 -1 1 1
 1 4 1 -1 3
 -1 -1 1 3 -1

<OUTPUT>
5

*/

C++: #include <iostream>

using namespace std;
#define MAX 100
#define NIL (1<<21)
#define WHITE 0
#define GRAY 1
#define BLACK 2
int G[MAX][MAX];//图的邻接矩阵
int color[MAX];//节点状态
int n;
int d[MAX];//最小权值
int p[MAX];//父亲节点
void prim(){
    for(int i=0;i<n;i++){
        color[i] = WHITE;
        d[i] = NIL;
        p[i] = -1;
    }
    d[0] = 0;
    while(1){
        int u;
        int mincost = NIL;
        for(int i=0;i<n;i++){//循环找到下一个要加入到T中的
            if(color[i]!=BLACK&&d[i]<mincost){
                mincost = d[i];
                u = i;
            }
        }
        if(mincost==NIL){
            break;
        }
        color[u] = BLACK;
        for(int v=0;v<n;v++){//给相邻顶点的d赋值（与之前的相比较，找到最小的）
            if(color[v]!=BLACK&&G[u][v]!=-1){
                if(G[u][v]<d[v]){
                    d[v] = G[u][v];
                    p[v] = u;
                    color[v] = GRAY;
                }
            }
        }
    }


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

    return 0;
}

C++: 
#include<iostream>

using namespace std;

static const int MAX = 100;
static const int INFTY = (1 << 21);
static const int WHITE = 0;
static const int GRAY = 1;
static const int BLACK = 2;


int n, M[MAX][MAX];;

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

  for (int i = 0; i < n; i++) {
    d[i] = INFTY;
    p[i] = -1;
    color[i] = WHITE;
  }

  d[0] = 0;

  while (1) {
    minv = INFTY;
    u = -1;
    for (int 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 (int v = 0; v < n; v++) {
      if (color[v] != BLACK && M[u][v] != INFTY) {
        if (d[v] > M[u][v]) {
          d[v] = M[u][v];
          p[v] = u;
          color[v] = GRAY;
        }
      }
    }
  }
  int sum = 0;
  for (int i = 0; i < n; i++) {
    if (p[i] != -1) {
      sum += M[i][p[i]];
    }
  }

  return sum;
}

int main() {
  cin >> n;

  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int e; cin >> e;
      M[i][j] = (e == -1) ? INFTY : e;
    }
  }

  cout << prim() << endl;

  return 0;
}

C++: #include <iostream>
using namespace std;

#define INF 1000000

int main(){
  long n;
  long G[100][100],d[100];
  long minc,mi,i;
  char flgs[i];
  long suma=0;
  
  cin >> n;
  for (i=0;i<n;i++){
    for (long j=0;j<n;j++){
      cin >> G[i][j];
      if (G[i][j]==-1) G[i][j]=INF;
    }
    d[i]=INF;
    flgs[i]=0;
  }

  d[0]=0;
  while (true){
    minc=INF;
    for (i=0;i<n;i++){
      if (flgs[i]!=1&&d[i]<minc){
	minc=d[i];
	mi=i;
      }
    }
    if (minc==INF) break;
    flgs[mi]=1;
    //for (i=0;i<n;i++)cout << flgs[i] << " ";
    for (i=0;i<n;i++){
      if (G[mi][i]==INF)continue;
      if (flgs[i]!=1&&d[i]>G[mi][i]){
	d[i]=G[mi][i];
	//cout << d[i];
      }
    }
  }
  
  for (i=0;i<n;i++){
    //cout << d[i]<<endl;
    suma+=d[i];
  }
  cout << suma<<endl;
  
  return 0;
}