Codeforces - 612D. The Union of k-Segments 排序+思维

1.题目描述：

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

2.题目大意：

给定n个区间，问你被覆盖至少k次的区间（两端连续区间可以合并）最少有多少个，并输出。

3.解题思路：

某个区间被覆盖至少k次，意味着在它前面的区间起点至少有k个且这些区间的终点不能出现在它前面。这样sort下，直接统计，遇到起点加一，终点减一，每k个一记录就行了。

4.AC代码：

#include <cstdio>
#include <algorithm>
#define INF 0x3f3f3f3f
#define maxn 10100000
using namespace std;

struct node
{
int p, val;
}num[maxn];

int cmp(node a, node b)
{
if (a.val != b.val)
return a.val < b.val;
return a.p > b.p;
}
int ans[maxn];
int main()
{
int n, k, top, cnt, flag;
while (scanf("%d%d", &n, &k) != EOF)
{
top = 0;
cnt = 0;
flag = 0;
for (int i = 0; i < n; i++)
{
int sta, ed;
scanf("%d%d", &sta, &ed);
num[top].val = sta;
num[top++].p = 1;
num[top].val = ed;
num[top++].p = 0;
}
sort(num, num + top, cmp);
for (int i = 0; i < top; i++)
{
if (num[i].p)
{
flag++;
if (flag == k)
ans[cnt++] = num[i].val;
}
else
{
if (flag == k)
ans[cnt++] = num[i].val;
flag--;
}
}
printf("%d\n", cnt / 2);
for (int i = 0; i < cnt; i += 2)
printf("%d %d\n", ans[i], ans[i + 1]);
}
return 0;
}