Range Queries with Sweep Line

Tutorial

Solution - Intersection Points

We can sweep from bottom to top (by the $y$ coordinate); storing two events for vertical segments (one for start and one for end) and one event for horizontal segments.

We can use a Binary Indexed Tree to store the number of active vertical segments for each $x$ coordinate.

Then, every time we encounter the start of a vertical segment, we increment the counter for $x$ in the BIT.

Similarly, we decrement the counter for $x$ every time we see the end of a vertical segment.

When we encounter a horizontal segment, we would query the number of active ranges in $[x_1, x_2]$ where $x_1$ is the lower $x$ coordinate and $x_2$ is the upper $x$ coordinate of the segment.

Our answer would be the sum of all the queries.

Copy
#include <bits/stdc++.h>
using namespace std;

int bit[2000005];

void update(int i, int x) {
	for (; i < 2000005; i += i & (-i)) bit[i] += x;
}
int query(int i) {
	int sum = 0;

Solution - Springboards

The problem boils down to having a data structure that supports the following operations:

1. Add a pair $(a,b)$.
2. For any $x$, query the minimum value of $b$ over all pairs satisfying $a\le x$.

The other module describes a simpler way of doing this, though it's probably more straightforward to do coordinate compression and use a min segtree. The latter approach is shown below.

Copy
/**
 * Description: 1D point update, range query where \texttt{comb} is
 * any associative operation. If $N=2^p$ then \texttt{seg[1]==query(0,N-1)}.
 * Time: O(\log N)
 * Source:
 * http://codeforces.com/blog/entry/18051
 * KACTL
 * Verification: SPOJ Fenwick
 */

Problems

Relating to LIS: