CodeForces 612D - The Union of k-Segments(模拟)

D. The Union of k-Segments

time limit per test4 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

题意:

给出n个线段和k,然后看在数轴上哪些区间的线段覆盖度大于等于k.

解题思路:

如果一个区间的覆盖度是k,那么他前面的线段起点必然大于等于k.

记录下每个线段的起点和终点.然后对所有点进行排序,如果遇到的是起点,那么cnt++,否则cnt–.

当cnt == k的时候,是这个区间的起点,再次遇到cnt == k同时该点是线段的终点时,这个点就是这个区间的终点.

AC代码:

#include<stdio.h>
#include<algorithm>
using namespace std;
const int maxn = 2000005;
struct node
{
int val;
int op;
};
node qdu[maxn];
int res[maxn];
bool cmp(node a,node b)
{
if(a.val != b.val)  return a.val < b.val;
else                return a.op > b.op;
}
int main()
{
int n;
int k;
scanf("%d%d",&n,&k);
int top = 0;
for(int i = 0;i < n;i++)
{
int l;
int r;
scanf("%d%d",&l,&r);
qdu[top].val = l;
qdu[top++].op = 1;
qdu[top].val = r;
qdu[top++].op = 0;
}
sort(qdu,qdu+top,cmp);
int have = 0;
int cnt = 0;
for(int i = 0;i < top;i++)
{
if(qdu[i].op)
{
have++;
if(have == k)    res[cnt++] = qdu[i].val;
}
else
{
if(have == k)    res[cnt++] = qdu[i].val;
have--;
}
}
printf("%d\n",cnt>>1);
for(int i = 0;i < cnt;i += 2)   printf("%d %d\n",res[i],res[i+1]);
return 0;
}