标签：1249D2 int Many Segments 30 li segments include ri
You are given n segments on the coordinate axis OX. Segments can intersect, lie inside each other and even coincide. The i-th segment is [li;ri](li≤ri) and it covers all integer points j such that li≤j≤ri.
The integer point is called bad if it is covered by strictly more than k segments.
Your task is to remove the minimum number of segments so that there are no bad points at all.
Input
The first line of the input contains two integers n and k (1≤k≤n≤2⋅105) — the number of segments and the maximum number of segments by which each integer point can be covered.
The next n lines contain segments. The i-th line contains two integers li and ri (1≤li≤ri≤2⋅105) — the endpoints of the i-th segment.
Output
In the first line print one integer m (0≤m≤n) — the minimum number of segments you need to remove so that there are no bad points.
In the second line print m distinct integers p1,p2,…,pm (1≤pi≤n) — indices of segments you remove in any order. If there are multiple answers, you can print any of them.
Examples
input
7 2
11 11
9 11
7 8
8 9
7 8
9 11
7 9
output
3
4 6 7
input
5 1
29 30
30 30
29 29
28 30
30 30
output
3
1 4 5
input
6 1
2 3
3 3
2 3
2 2
2 3
2 3
output
4
1 3 5 6
解题思路：
以区间左端点为头，右端点由小到大排序，如果端点相同，按标号存入。
从头开始遍历每个点，判断每个点上覆盖次数，大于k时删除最长的那个区间。
AC代码：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<set>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int N=2e5+10;
int n,k;
struct Node
{
int y;
int idx;
};
bool operator<(Node a,Node b)//排序
{
if(a.y!=b.y)
return a.y<b.y;
return a.idx<b.idx;
}
vector<Node> g[N];
vector<int> ans;
int main()
{
int x,y;
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
{
Node p;
scanf("%d%d",&x,&y);
p.y=y;
p.idx=i;
g[x].push_back(p);
}
set<Node> s;
for(int i=1;i<N;i++)
{
while(s.size()&&(*s.begin()).y<i)//没有被覆盖的区间删除
s.erase(*s.begin());
for(int j=0;j<g[i].size();j++)//存入可覆盖区间
s.insert(g[i][j]);
while(s.size()>k)//判断该点被覆盖次数是否超过 k
{
ans.push_back((*s.rbegin()).idx);//删除的区间存下
s.erase(*s.rbegin());
}
}
printf("%d\n",ans.size());
int len=ans.size();
for(int i=0;i<len;i++)
{
printf("%d%c",ans[i],i==len-1?'\n':' ');
}
return 0;
}
标签：1249D2,int,Many,Segments,30,li,segments,include,ri 来源： https://blog.csdn.net/weixin_43847416/article/details/102749985
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