Educational Codeforces Round 4-D. The Union of k-Segments

原题链接

D. The Union of k-Segments

time limit per test
4 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given n segments on the coordinate axis Ox and
the number k. The point is satisfied if it belongs to at least k segments.
Find the smallest (by the number of segments) set of segments on the coordinate axis Ox which contains all satisfied points
and no others.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 106)
— the number of segments and the value of k.

The next n lines contain two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109)
each — the endpoints of the i-th segment. The segments can degenerate and intersect each other. The segments are given in arbitrary order.

Output

First line contains integer m — the smallest number of segments.

Next m lines contain two integers aj, bj (aj ≤ bj)
— the ends of j-th segment in the answer. The segments should be listed in the order from left to right.

Examples

input

3 2
0 5
-3 2
3 8

output

2
0 2
3 5

input

3 2
0 5
-3 3
3 8

output

1
0 5

#include <bits/stdc++.h>
#define maxn 1000005
using namespace std;
typedef long long ll;

struct Node{
Node(){
}
Node(int a, int b){
l = a;
r = b;
}
int l, r;
}node[maxn];
vector<Node> v;
int n, k, d[maxn<<1], p[maxn<<1];
int main(){
// freopen("in.txt", "r", stdin);
scanf("%d%d", &n, &k);
for(int i = 0; i < n; i++){
scanf("%d%d", &node[i].l, &node[i].r);
d[2*i] = node[i].l;
d[2*i+1] = node[i].r;
}
sort(d, d+2*n);
int cnt = 2 * n;
for(int i = 0; i < n; i++){
int k1 = lower_bound(d, d+cnt, node[i].l) - d;
p[k1+1]++;
int k2 = upper_bound(d, d+cnt, node[i].r) - d;
p[k2]--;
}
for(int i = 1; i <= cnt; i++)
p[i] += p[i-1];
int l = -1, r = -1;
for(int i = 0; i <= cnt; i++){
if(l == -1 && p[i] >= k)
l = i;
if(l != -1 && p[i] < k){
r = i;
v.push_back(Node(d[l-1], d[r-1]));
l = -1;
}
}
printf("%d\n", v.size());
for(int i = 0; i < v.size(); i++)
printf("%d %d\n", v[i].l, v[i].r);
return 0;
}