D. The Union of k-Segments
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

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

题目大意：
给定n个区间，和k。给定的n个区间[l,r]会有重合，问有多少个区间，区间的重合个数大于k，并且输出区间
比如：n=2，k=2，给定俩区间[1,3],[2,4]，那么重合的部分就是[2,3]，并且重合的区间个数为2，大于等于k，所以输出[2,3]这个区间。
解题思路：
把每一个点封装成结构体，如果是左端点就flag=1，否则为-1。
然后把所有节点按从小到大排好序，从左到右扫描，遇见左端点就+1，否则减一（也就是value+=flag）
也就是用前缀和处理

如果觉得难理解，就手动模拟一下场景，我也是这样才明白的

#include <bits/stdc++.h>

using namespace std;

typedef struct segment
{
    int side,flag;
    bool operator <(const segment s)const
    {
        if(this->side != s.side)
            return this->side < s.side;
        else
            return this->flag >s.flag;
    }
}segment;
segment seg[2000005];
segment ans[2000005];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++)
    {
        scanf("%d %d",&seg[2*i].side, &seg[2*i+1].side);
        seg[2*i].flag=1;
        seg[2*i+1].flag=-1;
    }
    //共有2*n个点
    sort(seg,seg+2*n);
    int cnt=0,val=0;
    for(int i=0;i<2*n;i++)
    {
        val+=seg[i].flag;
//        printf("val=%d\n",val);
        if(k==val)
        {
//            printf("i=%d\n",i);
            ans[cnt++].side=i;
             while(k<=val && i<2*n)
            {
                i++;
                val+=seg[i].flag;
//                printf("k==%d\n",k);
            }
            ans[cnt++].side=i;
//            printf("aa==%d\n",ans[cnt-1].side);
        }
    }
//    printf("cnt=%d\n",cnt);
    printf("%d\n",cnt/2);
    for(int i=0;i<cnt/2;i++)
    {
//        printf("%d %d\n",ans[2*i].side  ,ans[2*i+1].side);
//        注意ans里面存的仅仅是下标,需要放到seg里面才能取值
        printf("%d %d\n",seg[ans[2*i].side].side,seg[ans[2*i+1].side].side);
    }
    return 0;
}
/*
2 2
1 3
2 4
*/