APIO 2019 - Bridges

Author: Andi Qu

The main idea in this problem is to use square-root decomposition on queries. For convenience, call type 1 queries updates and type 2 queries calculations.

First, split the queries into blocks of about N\sqrt N queries. In each block, there are O(N)\mathcal{O}(\sqrt N) updates or calculations. For each block:

  • Split the bridges into two groups: changed and unchanged.
  • If we sort the calculations and unchanged bridges in decreasing order of weight, we can simply use DSU to find which nodes are connected from those bridges alone.
    • These connected nodes are constant for all calculations in the current block
  • To handle the updates:
    • Iterate over the queries in the current block (without sorting)
    • If the query is an update, simply update the bridge's weight
    • If the query is a calculation, iterate through each changed bridge and connect the nodes if the weight limit is above the query's weight limit
      • This works because this means the answer for the current query is dependent only on previous updates
      • The key thing here is that we need a way to roll back DSU unions, since the set of "good" bridges may differ from query to query
      • To achieve this, we simply use DSU with path balancing only and keep a stack of previous DSU operations


Time Complexity: O((Q+M)logNQ)\mathcal{O}((Q + M) \log N \sqrt Q )

However, it is possible to remove the log factor as mentioned in this comment.


#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;
const int B = 1000;
int n, m, q;
stack<int> stck;

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