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

D. The Union of k-Segments

 

You re 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.

Sample test(s)

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

 题意：给你一n条线段，一个k,问你有k跳线段以上相交的线段有多少

题解：按照 左右端点区分，x,y一起排序，找增加至k个左端点加入左答案，减少到k-1个端点加入右答案，注意一个点的线段

//meek
///#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <sstream>
#include <queue>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define fi first
#define se second
#define MP make_pair
inline ll read()
{
    ll x=,f=;
    char ch=getchar();
    while(ch<''||ch>'')
    {
        if(ch=='-')f=-;
        ch=getchar();
    }
    while(ch>=''&&ch<='')
    {
        x=x*+ch-'';
        ch=getchar();
    }
    return x*f;
}
//****************************************
 
const int N=+;
const ll INF = 1ll<<;
const int inf = ;
const int mod= ;
 
int n,k,x,y,t[N];
pair<int,int>  Line[N];
vector<int >ans1,ans2;
int main() {
    scanf("%d%d",&n,&k);
    int cnt = ;
    for(int i=;i<=n;i++) {
        scanf("%d%d",&x,&y);
        Line[cnt++] = MP(x,-);
        Line[cnt++] = MP(y,);
    }
    sort(Line+,Line+cnt);
    for(int i=;i<cnt;i++)
    {
        t[i] = t[i-] - Line[i].se;
        if(t[i]==k&&t[i-]==k-)
            ans1.pb(Line[i].fi);
    }
    memset(t,,sizeof(t));
    for(int i=;i<cnt;i++)
    {
        t[i] = t[i-] - Line[i].se;
        if(t[i]==k-&&t[i-]==k)
            ans2.pb(Line[i].fi);
    }
    if(ans1.size()!=ans2.size()) ans2.pb(Line[cnt-].fi);
    cout<<ans1.size()<<endl;
    for(int i=;i<ans1.size();i++)
        cout<<ans1[i]<<" "<<ans2[i]<<endl;
    return ;
}

代码