F - Intervals

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本文介绍了一种算法，旨在解决如何最小化删除区间数量的问题，确保剩下的区间不会出现三个及以上的重叠情况。通过按区间右端点排序并跟踪重叠状态，实现高效求解。

Chiaki has n intervals and the i-th of them is [l_i, r_i]. She wants to delete some intervals so that there does not exist three intervals a, b and c such that aintersects with b, b intersects with c and c intersects with a.

Chiaki is interested in the minimum number of intervals which need to be deleted.

Note that interval a intersects with interval b if there exists a real number x such that l_a ≤ x ≤ r_a and l_b ≤ x ≤ r_b.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first line contains an integer n (1 ≤ n ≤ 50000) -- the number of intervals.

Each of the following n lines contains two integers l_i and r_i (1 ≤ l_i < r_i ≤ 10⁹) denoting the i-th interval. Note that for every 1 ≤ i < j ≤ n, l_i ≠ l_j or r_i ≠r_j.

It is guaranteed that the sum of all n does not exceed 500000.

Output

For each test case, output an integer m denoting the minimum number of deletions. Then in the next line, output m integers in increasing order denoting the index of the intervals to be deleted. If m equals to 0, you should output an empty line in the second line.

Sample Input

Sample Output

4
3 5 7 10

Hint

题意：问至少删除多少个区间才能保证剩下来的区间不会有三个及其以上的重叠。

思路:先将区间按右端点从小到大排序，记录一下重叠了两次的最右端位置right，若后面的区间的左端点小于它，则删除这个区间，若大于它，则将right更新=min（此区间的L，上一个区间的R）.

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
    int l,r,index;
}inter[50005];
bool cmp(node a,node b){
    if(a.r==b.r) return a.l<b.l;
    return a.r<b.r;
}
int sum,ans[50005];
int main(){
    int t;scanf("%d",&t);
    while(t--){
        int n;scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%d%d",&inter[i].l,&inter[i].r),inter[i].index=i+1;
        sort(inter,inter+n,cmp);
        sum=0;int right=0;node last=inter[0];
        for(int i=1;i<n;i++){
            if(inter[i].l<=right) {ans[sum++]=inter[i].index;continue;}
            if(inter[i].l<=last.r) right=last.r;
            last=inter[i];
        }
        sort(ans,ans+sum);
        printf("%d\n",sum);
        for(int i=0;i<sum-1;i++) printf("%d ",ans[i]);
        printf("%d\n",ans[sum-1]);
    }
}