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

D. The Union of k-Segments

题目连接：

http://www.codeforces.com/contest/612/problem/D

Description

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.

Sample Input

3 2

0 5

-3 2

3 8

Sample Output

2

0 2

3 5

Hint

题意

给你一堆区间，然后让你把覆盖k次及k次以上的区间都输出出来

题解：

直接暴力扫分界点就好了

分界点是正向覆盖k次的就加进左端点，是反向，就加进右端点，然后输出就好了

代码

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

#define maxn 3000006
pair<int,int> Line[maxn];
int tot = 1;
int t[maxn];
int main()
{
    int n,k;scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        Line[tot++]=make_pair(x,-1);
        Line[tot++]=make_pair(y,1);
    }
    sort(Line+1,Line+tot);
    int flag1 = 0,flag2 = 0;
    vector<int> ans1;
    vector<int> ans2;
    for(int i=1;i<tot;i++)
    {
        t[i] = t[i-1] - Line[i].second;
        if(t[i]==k&&t[i-1]==k-1)
            ans1.push_back(Line[i].first);
    }
    memset(t,0,sizeof(t));
    for(int i=1;i<tot;i++)
    {
        t[i] = t[i-1] - Line[i].second;

        if(t[i]==k-1&&t[i-1]==k)
            ans2.push_back(Line[i].first);
    }
    if(ans1.size()!=ans2.size())
        ans2.push_back(Line[tot-1].first);
    cout<<ans1.size()<<endl;
    for(int i=0;i<ans1.size();i++)
        cout<<ans1[i]<<" "<<ans2[i]<<endl;
}