APIO 2015 - Palembang Bridges

Time Complexity: $\mathcal O(N \log N)$

Since people only cross bridges when their two buildings are on opposite sides of the river, assume without loss of generality that all the people must cross a bridge.

$K = 1$

If the bridge is at position $x$, then the total cost would be $\sum_{i = 1}^N (|S_i - x| + |T_i - x|)$. Clearly, this value is minimized when $x$ is the median of all $S_i$ and $T_i$.

We can find the median simply by sorting all the numbers in the input.

$K = 2$

Since each person crosses exactly one bridge, which one would person $i$ choose? The answer is that they'd choose the bridge that is closest to $\frac{S_i + T_i}{2}$.

This means that if we sort the people by $S_i + T_i$, then we can split them into a prefix where people choose bridge 1 and a suffix where people choose bridge 2. Notice how we can use our solution for $K = 1$ to compute the answer for each bridge.

Now we can simply try out all the places to split the people into prefix and suffix. To maintain a sliding median, we can use two std::priority_queues. (For more details, see CSES Sliding Median)

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#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;

bool cmp(pair<int, int> a, pair<int, int> b) {
	return a.first + a.second < b.first + b.second;
}

priority_queue<int> lpq;