Overview

This is a problem of finding the minimum bridge construction cost required to connect all \(N\) islands. This is a classic Minimum Spanning Tree (MST) problem.

Analysis

Key Observations

1. Reformulation as Graph Theory: If we think of islands as vertices and bridges as edges, “all islands can travel between each other” means “the graph is connected.”

2. Number of Edges Required: To connect \(N\) vertices, at least \(N-1\) edges are needed. Also, \(N-1\) edges are sufficient (forming a tree structure).

3. Cost Minimization: When selecting \(N-1\) edges, we want to minimize the total cost → This is the minimum spanning tree problem.

Problems with the Naive Approach

If we try all combinations of edges, the number of ways to choose \(N-1\) edges from \(M\) edges is \(\binom{M}{N-1}\), which grows exponentially and results in TLE.

Solution

Using Kruskal’s algorithm, we can construct the minimum spanning tree by simply sorting edges by cost and greedily selecting them.

Algorithm

We use Kruskal’s Algorithm.

Procedure

1. Sort all edges in ascending order by cost
2. Prepare a Union-Find (disjoint set data structure)
3. Examine edges in order from lowest cost:
   • If the two endpoints of the edge belong to different connected components, adopt the edge and merge the components
   • If they belong to the same connected component, skip the edge as adopting it would create a cycle
4. Stop when \(N-1\) edges have been adopted
5. If ultimately fewer than \(N-1\) edges can be adopted, the graph cannot be connected, so output -1

Concrete Example

For \(N=4\), with edges \((1,2,10), (2,3,5), (3,4,8), (1,4,20), (2,4,15)\):

1. After sorting: \((2,3,5), (3,4,8), (1,2,10), (2,4,15), (1,4,20)\)
2. Adopt \((2,3,5)\) → {2,3} are connected
3. Adopt \((3,4,8)\) → {2,3,4} are connected
4. Adopt \((1,2,10)\) → {1,2,3,4} are connected
5. 3 edges adopted! Total cost = \(5 + 8 + 10 = 23\)

Complexity

• Time Complexity: \(O(M \log M)\)

  • Sorting edges: \(O(M \log M)\)
  • Union-Find operations (with path compression + union by rank): Each operation is nearly \(O(1)\), total \(O(M \cdot \alpha(N))\) ≈ \(O(M)\)

• Space Complexity: \(O(N + M)\)

  • Arrays for Union-Find: \(O(N)\)
  • Edge list: \(O(M)\)

Implementation Notes

1. Union-Find Implementation: By combining path compression and union by rank, operations can be performed in nearly constant time.

2. Special Case \(N = 1\): If there is only one island, no bridges are needed and the cost is \(0\).

3. Connectivity Check: This can be determined by whether the number of adopted edges becomes \(N-1\).

4. Overflow Prevention: The maximum cost is \(10^9\), and the maximum number of edges is \(N-1 \approx 10^5\), so the total cost is at most about \(10^{14}\). In Python, there is no need to worry about integer overflow.

5. Fast Input: By using sys.stdin.read() to read all input at once, large amounts of input can be processed quickly.

   Source Code

import sys
from typing import List, Tuple

def find(parent: List[int], x: int) -> int:
    if parent[x] != x:
        parent[x] = find(parent, parent[x])
    return parent[x]

def union(parent: List[int], rank: List[int], x: int, y: int) -> bool:
    px, py = find(parent, x), find(parent, y)
    if px == py:
        return False
    if rank[px] < rank[py]:
        px, py = py, px
    parent[py] = px
    if rank[px] == rank[py]:
        rank[px] += 1
    return True

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    
    if N == 1:
        print(0)
        return
    
    edges: List[Tuple[int, int, int]] = []
    for _ in range(M):
        u = int(input_data[idx]); idx += 1
        v = int(input_data[idx]); idx += 1
        c = int(input_data[idx]); idx += 1
        edges.append((c, u, v))
    
    # Sort edges by cost (Kruskal's algorithm)
    edges.sort()
    
    parent = list(range(N + 1))
    rank = [0] * (N + 1)
    
    total_cost = 0
    edge_count = 0
    
    for c, u, v in edges:
        if union(parent, rank, u, v):
            total_cost += c
            edge_count += 1
            if edge_count == N - 1:
                break
    
    if edge_count == N - 1:
        print(total_cost)
    else:
        print(-1)

if __name__ == "__main__":
    solve()

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