Problem:
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Minimum Spanning Tree

For a given weighted graph $G = (V, E)$, find the minimum spanning tree (MST) of $G$ and print total weight of edges belong to the MST.

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, a $n \times n$ adjacency matrix $A$ which represents $G$ is given.
$a_{ij}$ denotes the weight of edge connecting vertex $i$ and vertex $j$. If there is no edge between $i$ and $j$, $a_{ij}$ is given by -1.

Output

Print the total weight of the minimum spanning tree of $G$.

Constraints

$1 \leq n \leq 100$

$0 \leq a_{ij} \leq 2,000$ (if $a_{ij} \neq -1$)

$a_{ij} = a_{ji}$

$G$ is a connected graph

Sample Input 1

5
-1 2 3 1 -1
2 -1 -1 4 -1
3 -1 -1 1 1
1 4 1 -1 3
-1 -1 1 3 -1

Sample Output 1

5

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Python3 (judge verdict: Accepted)

import sys
import heapq

def main():
    n = int(input())
    next = []
    for _ in range(n):
        l = list(map(int, input().split()))
        next.append([(c, i) for i, c in enumerate(l) if c != -1])
    visited = [0] * n
    connection = 0
    q = [(0, 0)]    # (cost, v)
    heapq.heapify(q)
    ans = 0
    while len(q):
        cost, v = heapq.heappop(q)
        if visited[v]:
            continue

        # そのノードと繋げる
        visited[v] = 1
        connection += 1
        ans += cost
        
        # 新たに繋げたノードから行けるところをエンキュー
        for i in next[v]:
            heapq.heappush(q, i)
        
        # 全部のノードが繋がったら終了
        if connection == n:
            break
    print(ans)
    return

if __name__ == '__main__':
    main()