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++: // Graph II - Mini#include<iostream>
#include<algorithm>
#include<vector>
#include<functional>
using namespace std;

#if 0
所用時間: 1h

感想: 思ったより実装すべきコード量が多かった。DisjointSetは今後も使う機会が多そうなので、クラス化してみた。
Pの配列を動的に確保するよう工夫した、良い復習になった。

#endif

int N, buff, M = 0, SUM = 0, A[101][101], cost[5000], dpt[5000], dst[5000];//A[Y][X]
vector<pair<int, int> > edges;

class disjointSet {
private:
    int* P;
    
public:
    disjointSet(){
        P = new int[100000];
    }
    disjointSet(int N){     
        P = new int[N];
    }
    ~disjointSet(){
        delete[] P;
    }

    void init(int N){ //木構造の初期化
        for(int i = 0; i <= N; ++i){
            P[i] = i;
        }
    }

    int root(int a){ //根の参照
        if(P[a] == a) return a;
        return (P[a] = root(P[a]));
    }

    bool is_same_set(int a, int b){ //同じ根か?
        return root(a) == root(b);
    }

    void unite(int a, int b){
        P[root(a)] = root(b); //aのrootをbの子にする
    }
};


void preprocess(){//辺の前処理
  for(int r = 0; r < N; ++r){
    for(int c = r+1; c < N; ++c){
      if (A[r][c] > -1) {
        cost[M] = A[r][c];
        dpt[M] = r;
        dst[M] = c;
        M++;
      }
    }
  }

  //for(int i = 0; i < M; ++i) cout << cost[i] << endl;

  for(int i = 0; i < M; ++i){
    edges.push_back(make_pair(cost[i], i));
  }
  for(int i = 0; i < M; ++i) cout << edges[i].first << " " << edges[i].second << endl;

  sort(edges.begin(), edges.end());

  for(int i = 0; i < M; ++i) cout << edges[i].first << " " << edges[i].second << endl;

}

void solve(){
  disjointSet Union(100000);

  int ans = 0;
  preprocess();

  Union.init(N);

  for(int i = 0; i < M; ++i){
    int index = edges[i].second;
    if(Union.is_same_set(dpt[index], dst[index])) continue;
    else {
      Union.unite(dpt[index], dst[index]);
      cout << "unite " << dpt[index] << " and " << dst[index] << endl;
      ans += cost[index];
    }
  }
  cout << ans << endl;
}

int main(){

cin >> N;

for(int i = 0; i < N; ++i){
  for(int j = 0; j < N; ++j){
    cin >> buff;
    A[i][j] = buff;
    }
  }

  //行列の表示
  // cout << "Matrix:" << endl;
  // for(int row=0; row < N; ++row){
  //   for(int col = 0; col < N; ++col){
  //     cout << A[row][col];
  //   }
  //   cout << endl;
  // }

  solve();
}
mum Spanning Tree
// 問題URL: http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_12_A&lang=jp
//
// 使うライブラリも多く、難しく感じた...
#include <iostream>
#include <algorithm>

using namespace std;

int n, buff, cost[101][101]={}, P[101], M = 0, ans = 0;
pair<int, int> edges[10010], vertices[10010];

void init(int n){
  for(int i = 0; i < n; ++i){
    P[i] = i;
  }
}

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

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

void unite(int x, int y){
  P[root(x)] = root(y);
}

int main(){
  cin >> n;

  for(int row = 0; row < n; ++row){
    for(int col = 0; col < n; ++col){
      cin >> buff;
      cost[row][col] = buff; //隣接行列の読込
    }
  }

  for(int row = 0; row < n; ++row){
    for(int col = 0; col < row; ++col){
      if(cost[row][col]>0){
        edges[M].first = cost[row][col];
        edges[M].second = M;
        // cout << M << ": (" << row << ", " << col << ") = " << edges[M].first << endl;
        vertices[M].first = row, vertices[M].second = col;
        M++; //辺の数を管理、idとして利用
      }
    }
  }

  init(n);

  sort(edges, edges+M); //重みの小さい順に整列

  for(int i = 0; i <= M; ++i){
    int index = edges[i].second; //辺のidを取り出す
    if(!is_same_set(vertices[index].first, vertices[index].second)){ //辺の両端が繋がっていなければ
      unite(vertices[index].first, vertices[index].second); //両端を繋げる
      // cout << "unite " << vertices[index].first << " and " << vertices[index].second << endl;
      ans += edges[i].first; //コストを加える
    }
  }
  cout << ans << endl;
}

C++: //include
//------------------------------------------
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <limits>
using namespace std;

//repetition
//------------------------------------------
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORR(i,a,b) for(int i=(a);i>(b);i--)
#define REP(i,n)  FOR(i,0,n)

//conversion
//------------------------------------------
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

//debug
//-------------------------------------------
#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
#define INF std::numeric_limits<int>::max()

//typedef
//-------------------------------------------
typedef long long ll;

int n;
int g[101][101];

int prim(int used, int sum) {
    if (used == 0) return sum;

    int tmp = INF;
    int use = -1;
    REP(i, n) {
        if ((used>>i)&1) {
            continue;
        }
        REP(j, n) {
            if (((used>>j)&1) && g[i][j] < tmp) {
                tmp = g[i][j];
                use = j;
            }
        }
    }
    if (tmp == INF) return sum;
    return prim(used&(~(1<<use)), sum + tmp);
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    REP(i, 101) REP(j, 101) g[i][j] = INF;
    cin >> n;
    REP(i, n) REP(j, n) {
        cin >> g[i][j];
        if (g[i][j] < 0) g[i][j] = INF;
    }
    int used = (1<<n) - 1;
    cout << prim(used&(~1), 0) << endl;

    return 0;
}