Educational Codeforces Round 4 612D The Union of k-Segments(stl)

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.

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

题目链接:点击打开链接

给出n段线段, 有重合的部分, 输出所有线段, 要求有k部分重合.

对线段的左端点标记-1, 右端点标记1, 排序后进行遍历, sum记录重合的线段数, 记录到k - 1则标记左端点, 记录到k则push到ans.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
int n, k, x, sum;
std::vector<pair<int, int> > v, ans;
int main(int argc, char const *argv[])
{
scanf("%d%d", &n, &k);
for(int i = 0; i < n; ++i) {
int l, r;
scanf("%d%d", &l, &r);
v.push_back(make_pair(l, -1));
v.push_back(make_pair(r, 1));
}
sort(v.begin(), v.end());
int len = v.size();
for(int i = 0; i < len; ++i) {
if(sum == k - 1 && v[i].second == -1) x = v[i].first;
if(sum == k && v[i].second == 1) ans.push_back(make_pair(x, v[i].first));
sum -= v[i].second;
}
len = ans.size();
cout << len << endl;
for(int i = 0; i < len; ++i)
printf("%d %d\n", ans[i].first, ans[i].second);
return 0;
}