The only difference between easy and hard versions is constraints.
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≤200) — 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≤200) — 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.
题意： 一个点被超过k个线段覆盖就被称为坏点，给出n条线段，求至少删除几个线段不会出现坏点。
思路： 如果一个点被覆盖超过k次，首先删除区间右端点靠左的，这样对以后的点的影响会最小，所以按照右端点从小到大的方式对所有线段进行排序，然后进行贪心找到所有需要删除的线段。
代码：
#include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> using namespace std; int n,m; int a,b,p[1010],ans[1010]; struct node { int x,y,id; }s[1010]; bool cmp(node a,node b) { if(a.y==b.y) return a.x<b.x; else return a.y<b.y; } int main() { while(~scanf("%d%d",&n,&m)) { memset(p,0,sizeof p); memset(ans,0,sizeof ans); for(int i=0;i<n;i++) { scanf("%d%d",&a,&b); s[i].x=a; s[i].y=b; s[i].id=i+1; } sort(s,s+n,cmp); int k=0; for(int i=0;i<n;i++) { for(int j=s[i].x;j<=s[i].y;j++) { ans[j]++; if(ans[j]>m) { p[k++]=s[i].id; break; } } } printf("%d\n",k); for(int i=0;i<k;i++) printf("%d ",p[i]); printf("\n"); } return 0; }