Points in Segments
Hazrat Ali
You are given a set of n segments on the axis Ox, each segment has integer endpoints between 1 and m inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers li and ri (1≤li≤ri≤m) — coordinates of the left and of the right endpoints.
Consider all integer points between 1 and m inclusive. Your task is to print all such points that don't belong to any segment. The point x belongs to the segment [l;r] if and only if l≤x≤r.
The first line of the input contains two integers n and m (1≤n,m≤100) — the number of segments and the upper bound for coordinates.
The next n lines contain two integers each li and ri (1≤li≤ri≤m) — the endpoints of the i-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that li=ri, i.e. a segment can degenerate to a point.
In the first line print one integer k — the number of points that don't belong to any segment.
In the second line print exactly k integers in any order — the points that don't belong to any segment. All points you print should be distinct.
If there are no such points at all, print a single integer 0 in the first line and either leave the second line empty or do not print it at all.
3 5
2 2
1 2
5 5
2
3 4
1 7
1 7
0
Solution
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
vector<int> s(m);
for (int i = 0; i < n; i++) {
int l, r;
cin >> l >> r;
for (int j = l - 1; j < r; j++) {
s[j]++;
}
}
int cnt = 0;
for (int i = 0; i < m; i++) {
if (s[i] == 0) {
cnt++;
}
}
cout << cnt << "\n";
for (int i = 0; i < m; i++) {
if (s[i] == 0) {
cout << i + 1 << " ";
}
}
cout << "\n";
return 0;
}