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>
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]];
	}
} 

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();
	
	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]; 
    // Tに属する頂点とV-Tに属する頂点をつなぐ辺の中で、重みが最小の辺の重みを記録
    // MSTにおける頂点vの親を記録
    // 訪問状態を記録

    for(int i=0; i<n; i++){ // 全ての頂点に関して、colorをwhite、dをinftyに初期化
        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++){ // 最小の辺を探索
            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){ // 未訪問かつuからvに辺があるとき
                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(int argc, char const *argv[])
{
    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>
#include <utility>
#include <algorithm>
using namespace std;

int p[100];
pair<int, int> a[10000];

int root( int m ) {
	if ( p[m] != m )
		p[m] = root( p[m] );
	return p[m];
}

void unit( int x, int y ) {
	p[root( y )] = root( x );
}

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

int main() {
	int n, ai, pn, ans;
	cin >> n;
	for ( int i = 0; i < n * n; i++ ) {
		cin >> ai;
		if ( ai == -1 || i % n <= i / n )
			ai = 2001;
		a[i] = make_pair( ai, i );
	}
	sort( a, a + ( n * n ) );
	pn = 0;
	ans = 0;
	for ( int i = 0; i < n; i++ )
		p[i] = i;
	for ( int i = 0; i < n * n; i++ ) {
		if ( ! same( a[i].second % n, a[i].second / n ) ) {
			ans += a[i].first;
			pn++;
			unit( a[i].second % n, a[i].second / n );
			if ( pn >= n - 1 ) break;
		}
	}
	cout << ans << endl;
}