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 <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <climits>


#define MAX 100


using namespace std;


int main()
{
    vector< vector<int> > adj(MAX + 1, vector<int>(MAX, INT_MAX));
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            int cost;
            cin >> cost;
            if (cost >= 0)
                adj[i][j] = cost;
        }
    }

    vector<int> parent(MAX);
    vector<int> in_tree(MAX, false);
    vector<int> min_dists(MAX, INT_MAX);
    int tree_count = 0;

    int acc = 0;
    min_dists[0] = 0;
    for (int tc = 0; tc < n; tc++) {
        int min_cost = INT_MAX;
        int min_v = -1;
        for (int i = 0; i < n; i++) {
            if (min_cost > min_dists[i] && !in_tree[i]) {
                min_cost = min_dists[i];
                min_v = i;
            }
        }

        acc += min_cost;
        in_tree[min_v] = true;

        for (int i = 0; i < n; i++) {
            if (min_dists[i] > adj[min_v][i] && !in_tree[i]) {
                min_dists[i] = adj[min_v][i];
                parent[i] = min_v;
            }
        }
    }

    cout << acc << 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;
  bool flgs[i];
  long suma=0;
  
  cin >> n;
  for (i=0;i<n;i++){
    for (int j=0;j<n;j++){
      cin >> G[i][j];
      if (G[i][j]==-1) G[i][j]=INF;
    }
    d[i]=INF;
    flgs[i]=false;
  }

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

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

#define ll long long int
#define P pair<ll,ll>

const ll INF = 10000000000;
struct edge{ll to,cost;};

int v;
vector<edge> g[100000];
ll d[100000];

int prim(){
    priority_queue<P,vector<P>,greater<P> > que;
    fill(d,d+v,INF);
    int res = 0;
    d[0] = 0;
    que.push(P(0,0));

    while(!que.empty()){
        P p = que.top(); que.pop();
        ll k = p.second;
        if(d[k] < p.first){
            continue;
        }
        res += d[k];
        for(int i=0; i < g[k].size(); i++){
            edge e = g[k][i];
            if(d[e.to] > /*d[k]*/ + e.cost){
                d[e.to] = /*d[k] */+ e.cost;
                que.push(P(d[e.to],e.to));
            }
        }
    }
    return res;
}

int main() {

    int s;

    cin >> v;

    for(int i=0; i<v; i++){
        for(int j=0;j<v;j++){
            cin >> s;
            if(s!=-1){
                edge ed;
                ed.to = j;
                ed.cost = s;
                g[i].push_back(ed);
            }
        }
    }

    /*for(int i=0; i<v; i++){
        for(int j=0;j<g[i].size();j++){
            cout << g[i][j].to << "," << g[i][j].cost << " ";
        }
        cout << endl;
    }*/

    cout << prim() << endl;

    return 0;
}

C++: #include <iostream>
#include <algorithm>
#include <vector>
#include <deque>
#include <queue>
#include <list>
#include <stack>
#include <string>
#include <tuple>
#include <functional>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <map>
#include <random>
#include <cstdlib>
#define INT_MAX 2147483647
#define INT_MIN -2147483646
#define Loop(i, n) for(int i = 0; i < (int)n; i++)
#define Loop1(i, n) for(int i = 1; i <= (int)n; i++)
#define Loopr(i, n) for(int i = (int)n - 1; i >= 0; i--)
#define Loopr1(i, n) for(int i = (int)n; i >= 1; i--)
using namespace std;
typedef long long int ll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;

static const int N = 105;

struct vertex {
	int id;
	int p;
	int d;
	bool operator<(const vertex & another) const {
		return (d != another.d ? d>another.d : id>another.id);
	}
};

int main() {
	vvi mx(N, vi(N, -1));
	int n;
	cin >> n;
	Loop(i, n) {
		Loop(j, n) {
			int a;
			cin >> a;
			mx[i][j] = a;
		}
	}
	//
	vi color(N, 0); //0,1
	priority_queue<vertex> q;
	q.push({ 0,-1,0});
	int ret = 0;
	while(q.size() > 0) {//?
		vertex v = q.top();
		q.pop();
		if (color[v.id] == 1) continue;
		color[v.id] = 1;
		ret += v.d;
		Loop(j, n) {
			if (mx[v.id][j] != -1 && color[j] != 1){
				q.push({ j, v.id,mx[v.id][j] });
			}
		}
	}
	cout << ret << endl;
	return 0;
}

C++: #if 0
time: 20min
using pair to store the edges' number, making it easier to find their nodes.					
#endif
#include <iostream>
#include <utility>
#include <algorithm>
using namespace std;

int m,n,x,ans;
int P[100+10],edge1[10000+10],edge2[10000+10];


int root(int a){
	if (P[a] == a) return a;
	return (P[a] = root(P[a]));
}

bool same(int a, int b){
	return root(a)==root(b);
}
	
void unite(int a, int b){
	P[root(a)] = root(b);
}

int main() {
	pair<int,int> edges[10000+10];
	cin >> n;
	
	for (int i=1; i<=n; i++){
		P[i] = i;
	}
	
	m = 0;
	for (int i=1; i<=n; i++){
		for (int j=1; j<=n; j++){
			cin >> x;
			if (x>-1){
				edge1[m] = i;
				edge2[m] = j;
				edges[m] = make_pair(x, m);
				m+=1;
			}
		}
	}
	
	sort(edges, edges+m);
	n=1;
	for (int i=0; i<m; i++){
		if (n>=m) break;
		if (!same(edge1[edges[i].second], edge2[edges[i].second])){
			unite(edge1[edges[i].second], edge2[edges[i].second]);
			ans += edges[i].first;
			n+=1;
		}
	}
	
	cout << ans << endl;
	
}