Go: package main

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

const BIG_COST = 3000

const (
	ST_NOT_YET = iota
	ST_JOINED
)

func main() {
	sc := bufio.NewScanner(os.Stdin)
	sc.Split(bufio.ScanWords)

	nNodes := scanInt(sc)
	costs := make([][]int, nNodes)

	for r := 0; r < nNodes; r++ {
		costs[r] = make([]int, nNodes)
		for c := 0; c < nNodes; c++ {
			costs[r][c] = scanInt(sc)
		}
	}

	states := make([]int, nNodes)
	minCosts := make([]int, nNodes)

	for i := 1; i < nNodes; i++ {
		minCosts[i] = BIG_COST
	}

	for {
		var u int
		minCost := BIG_COST

		for i := 0; i < nNodes; i++ {
			if states[i] != ST_JOINED && minCosts[i] < minCost {
				minCost = minCosts[i]
				u = i
			}
		}

		// 全てのノードが MST に JOIN できたので終了。
		if minCost == BIG_COST {
			break
		}

		states[u] = ST_JOINED

		for v := 0; v < nNodes; v++ {
			cost := costs[u][v]

			// MST内の u と MST外の v との辺の重みをそれぞれ minCosts[v] として
			// 保存しておく。このうち辺の重みが最小となる v が次のループで MST に JOIN する。
			if states[v] != ST_JOINED && cost != -1 && cost < minCosts[v] {
				minCosts[v] = cost
			}
		}
	}

	sum := 0
	for _, cost := range minCosts {
		sum += cost
	}
	fmt.Println(sum)
}

func scanInt(sc *bufio.Scanner) int {
	sc.Scan()
	v, _ := strconv.Atoi(sc.Text())
	return v
}

Python: import heapq


class MinimumSpanningTree:
    """ Prim's algorithm: find minimum spanning tree
        Complexity: O(E log(V))
        used in GRL2A(AOJ)
    """

    def __init__(self, V, E, start=0, INF=10**9):
        """ V: the number of vertexes
            E: adjacency list (undirected graph)
            start: start vertex
            INF: Infinity cost
        """
        self.prim(V, E, start, INF)

    def prim(self, V, E, start=0, INF=10**9):
        used = [False] * V
        self.mincost = 0
        que = []
        heapq.heappush(que, (0, 0))
        while len(que) > 0:
            cost, v = heapq.heappop(que)
            if used[v]:
                continue
            used[v] = True
            self.mincost += cost
            for to, c in E[v]:
                heapq.heappush(que, (c, to))

    def minCost(self):
        return self.mincost

V = int(input())
edge = [[] for _ in range(V)]
for i in range(V):
    for j, d in enumerate(map(int, input().split())):
        if d != -1:
            edge[i].append((j, d))
mst = MinimumSpanningTree(V, edge)
print(mst.minCost())

Python: n = int(input())
m = []

for _ in range(n):
    tmp = [int(x) for x in input().split()]
    m.append(tmp)

edge_list = [] 

j = 1
for i in range(n-1):
    for k,x in enumerate(m[i][j:]):
        if x != -1:
            edge_list.append((x,i,k))
        else:
            pass
    j += 1
edge_list.sort()

root_list = [i for i in range(n)]

def root(x):
    path_to_root = []
    while root_list[x] != x:
        path_to_root.append(x)
        x = root_list[x]
    for node in path_to_root:
        root_list[node] = x
    return x

def is_same_root(x,y):
    return root(x) == root(y)

def unite(x,y):
    root_list[root(x)] = root(y)

cost = 0
for x in edge_list:
    w,p,q = x
    if is_same_root(p,q):
        pass
    else:
        cost += w
        unite(p,q)

print(cost)

Python: n = int(input())

a = []
for i in range(n):
    a.append(list(map(int, input().split())))

ans = 0
mincost = [2001] * n
flag = [0] * n
mincost[0] = 0

while 1:
    v = -1
    for j in range(n):
        if flag[j] != 1 and (v == -1 or mincost[j] < mincost[v]):
            v = j
    if v == -1:
        break
    flag[v] = 1
    ans += mincost[v]
    for j in range(n):
        if a[v][j] == -1:
            continue
        mincost[j] = min(mincost[j], a[v][j])
print(ans)

Python: import heapq
from collections import defaultdict
v_num = int(input())
edges = defaultdict(list)
for i in range(v_num):
    data = [int(n) for n in input().split(" ")[1:]]
    for j in range(i + 1, v_num):
        if data[j] != -1:
            edges[i].append([data[j], j])
            edges[j].append([data[j], i])
queue = []
heapq.heapify(queue)
for c in edges[0]:
    heapq.heappush(queue, c)
visited = {0}
answer = 0
while len(visited) < v_num:
    q = heapq.heappop(queue)
    weight, one = q[0], q[1]
    if one in visited:
        continue
    answer += weight
    for e in edges[one]:
        heapq.heappush(queue, e)
    visited |= {one}
print(answer)

Python: def prim(n):
    d[0] = 0
    while True:
        cost_min = 10**9
        u = -1
        for i in range(n):
            weight = d[i]
            if flag_list[i] < 2 and weight < cost_min:
                u = i
                cost_min = weight
        if u == -1:
            break
        flag_list[u] = 2
        for v in range(n):
            edge = m[u][v]
            if flag_list[v] < 2 and edge > -1:
                if d[v] > edge:
                    d[v] = edge
                    p[v] = u
                    flag_list[v] = 1
    cnt = 0
    for i in range(n):
        if p[i] != -1:
            cnt += m[i][p[i]]
    return cnt


if __name__ == "__main__":
    n = int(input())
    m = []
    for _ in range(n):
        in_list = list(map(int, input().split()))
        m.append(in_list)

    flag_list = [0]*n # 0: 未到達, 1: 処理中, 2: 完了
    d = [10**9]*n
    p = [-1]*n
    print(prim(n))

Python: def kruskal(n):
    cost = []
    tree = [{i} for i in range(n)]
    while edge:
        cmin = min(weight)
        index = weight.index(cmin)
        i, j = edge[index]
        del weight[index], edge[index]
        for t in tree:
            if {i, j} <= t:
                break
        else:
            cost.append(cmin)
            for n in range(len(tree)):
                if i in tree[n]:
                    tmp = tree[n]
                    del tree[n]
                    break
            for n in range(len(tree)):
                if j in tree[n]:
                    tree[n] |= tmp
                    break
    return sum(cost)

n = int(input())
edge = []
weight = []
for i in range(n):
    for j, c in enumerate(map(int, input().split())):
        if c != -1 and not {i, j} in weight:
            edge.append({i, j})
            weight.append(c)
print(kruskal(n))

Python: 
INFINITY = 10 ** 10
WHITE = 0
GRAY = 1
BLACK = 2


n = int(input())
M = [[INFINITY] * n for _ in range(n)]

for i in range(n):
    for j, weight in enumerate(map(int, input().split())):
        if weight == -1:
            weight = INFINITY

        M[i][j] = weight


def prim():
    color = [WHITE] * n
    d = [INFINITY] * n
    p = [None] * n

    d[0] = 0
    p[0] = -1

    while True:
        cost_min = INFINITY
        for i in range(n):
            if color[i] != BLACK and d[i] < cost_min:
                cost_min = d[i]
                top_now = i

        if cost_min == INFINITY:
            break
    
        color[top_now] = BLACK

        for v in range(n):
            if color[v] == BLACK or M[top_now][v] == INFINITY:
                continue

            if M[top_now][v] < d[v]:
                d[v] = M[top_now][v]
                p[v] = top_now
                color[v] = GRAY

    return sum(d)

cost_min = prim()
print(cost_min)

Python: n=int(input())
m = [[-1 for i in range(n)] for j in range(n)]
for i in range(n): #隣接の辺のコストを入力
  nums = list(map(int, input().split()))
  for j in range(n):
    m[i][j] = nums[j]

d=[0]+[10**9]*(n-1)
visited=[0]*n

while 1:
  u=-1
  for i in range(n):
    if not visited[i] and (u<0 or d[i]<d[u]):
      u = i
  if u < 0:
    break
  visited[u]=1
  for v in range(n):
    if not visited[v] and m[u][v]!=-1:
      if m[u][v]<d[v]:
        d[v]=m[u][v]

print(sum(d))