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++: #include<bits/stdc++.h>
using namespace std;
#define INF 10000000

int prim(int v, vector<vector<pair<int, int> > > adjlist){

    priority_queue<pair<int, int> >  wait;    // first := 候補辺のコスト   second　:= 行先ノード
    vector<int> result(v, INF);               // 頂点iと木の間の最小コスト
    int ans = 0, size = 0;                    // 木のコスト　頂点数

    //初期条件
    result[0] = 0;
    wait.push(make_pair(0, 0));

    while(!wait.empty() && size < v){

        //候補辺のうち最小のものを取り出す
        int newpoint = wait.top().second;
        int newcost = -wait.top().first;
        wait.pop();

        if(result[newpoint] < newcost) continue;

        //出てきた頂点を木に追加
        ans += newcost;
        result[newpoint] = -INF;
        size++;

        // 新しい点から出ている辺をすべてなめる
        for(int i = 0; i < adjlist[newpoint].size(); i++){
            int nextpoint = adjlist[newpoint][i].second;
            int nextcost = adjlist[newpoint][i].first;
            //より木に近くなった頂点があれば,　そこへの枝をqueueに追加
            if(result[nextpoint] > nextcost){
                result[nextpoint] = nextcost;
                wait.push(make_pair(-nextcost, nextpoint));
            }
        }
    }

    return ans;
}


int main(){

    int V; cin >> V;     // V:= 頂点の数
    vector<vector<pair<int, int> > > adjlist(V);      //first := コスト　second := 行先

    //入力
    for(int i = 0; i < V; i++){
        for(int j = 0; j < V; j++){
            int a; cin >> a;
            if(a == -1) continue;
            else{
                adjlist[i].push_back(make_pair(a, j));
            }
        }  
    }

    int ans = prim(V, adjlist);

    cout << ans << endl;
    return 0;
}

C++: #include "stdafx.h"
#include <iostream>

#define MAX 100000000;
using namespace std;

int main(void)
{
	int n, now = 0, flag[101] = {};
	long ans = 0, an[100][100], min[100];

	cin >> n;

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++)
			cin >> an[i][j];
		min[i] = MAX;
	}

	do {
		long key = , key_i;
		flag[now] = 1;
		flag[100] = 0;

		for (int i = 0; i < n; i++) {
			if (an[now][i] != -1 && an[now][i] < min[i] && flag[i] == 0)
				min[i] = an[now][i];
			if (flag[i] == 0 && min[i] < key) {
				key = min[i];
				key_i = i;
			}
			if (flag[i] == 1)
				flag[100]++;
		}

		now = key_i;
		ans += key;

	} while (flag[100] != n - 1);

	cout << ans << endl;

	return 0;
}

C++: #include <bits/stdc++.h>
using namespace std;
struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
#define P(x) cout << (x) << '\n'
#define p(x) cout << (x)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define vv(type, c, m, n, i) vector<vector<type>> c(m, vector<type>(n, i));
#define rep(i,a,n) for(int i=(a), i##_len=(n); i<i##_len; ++i)
#define rrep(i,a,n) for(int i=(a), i##_len=(n); i>i##_len; --i)
#define len(x) ((int)(x).size())
#define mp make_pair
#define eb emplace_back
typedef long long ll;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<long long> vll;
typedef vector<string> vs;
typedef vector<bool> vb;
constexpr int MAX = 101;
constexpr int INF = 1<<21;
enum Color { WHITE = 0, GRAY, BLACK };
int n;
vv(int, M, MAX, MAX, INF);
inline int prim() {
    vi d(n, INF), t(n, -1), color(n, WHITE);
    d[0] = 0;
    int minv, u;
    while(true) {
        minv = INF;
        u = -1;
        rep(i, 0, n) {
            if(minv > d[i] && color[i] != BLACK) {
                u = i;
                minv = d[i];
            }
        }
        if(u == -1) break;
        color[u] = BLACK;
        rep(v, 0, n) {
            if(color[v] != BLACK && d[v] > M[u][v]) {
                d[v] = M[u][v];
                t[v] = u;
                color[v] = GRAY;
            }
        }
    }
    return accumulate(all(d), 0);
}
int main() {
    cin >> n;
    int e;
    rep(i, 0, n) {
        rep(j, 0, n) {
            cin >> e;
            if(e != -1) M[i][j] = e;
        }
    }
    P(prim());
    return 0;
}