zoj 3953 三区间重合（贪心）

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解决一个算法问题，即如何最小化删除的区间数量以确保不存在三个互相交叠的区间。输入包括多个测试案例，每个案例中有一组区间，目标是确定并输出需要删除的区间及其索引。

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 a intersects 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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<string>
#include<map>
#include<set>
using namespace std;
const int N = 50000 + 100;
struct P
{
    int x,y,id;
    bool operator < (const P& t1)const
    {
        return y<t1.y;
    }
}a[N];
int b[N];
int vis[N];
int main()
{
    int T,n,i,j,end,ans;
    scanf("%d",&T);
    while(T--) {
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        for(i=1;i<=n;i++) {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].id=i;
        }
        sort(a+1,a+1+n);
        ans=end=0;
        P last = a[1];
        for(i=2;i<=n;i++) {
            if(a[i].x<=end) {
                vis[i]=1;
                continue;
            }
            if(a[i].x<=last.y) end = last.y;
            last = a[i];
        }
        for(i=1;i<=n;i++)
            if(vis[i]) b[ans++]=a[i].id;
        printf("%d\n",ans);
        sort(b,b+ans);
        for(i=0;i<ans;i++) printf("%d ",b[i]);
        printf("\n");
    }
    return 0;
}