ZOJ

Intervals

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Time Limit: 1 Second      Memory Limit: 65536 KB      Special Judge

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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

1112 54 73 96 111 1210 158 1713 1816 2014 2119 22

Sample Output

4
3 5 7 10
题目大意：
给你n个区间的左右端点，让你取走一些区间，使得任何三个区间都不两两相交。
思路：
贪心，首先将所有区间以左端点从小到大排序，然后每次判断当前的三个区间是否两两相交，如果两两相交则删除右端点值最大的那个区间，因为这样对后面的影响会最小。如果不相交则继续添加区间，每三个一判断。
#include<bits/stdc++.h>using namespace std;const int maxn = 50005;struct Intervals{    int l, r, id;}a[maxn], tmp[5];bool cmp1(Intervals p, Intervals q){    if(p.l == q.l)        return p.r < q.r;    return p.l < q.l;}bool cmp2(Intervals p, Intervals q){    return p.r > q.r;}bool isInterval(Intervals x, Intervals y, Intervals z){    return y.l <= x.r && z.l <= y.r && z.l <= x.r;}int ans[maxn], cnt;int main(){    int n, T;    scanf("%d", &T);    while(T--)    {        cnt = 0;        scanf("%d", &n);        for(int i = 1; i <= n; i++)        {            scanf("%d%d", &a[i].l, &a[i].r);            a[i].id = i;        }        sort(a+1, a+n+1, cmp1);        tmp[1] = a[1];        tmp[2] = a[2];        for(int i = 3; i <= n; i++)        {            tmp[3] = a[i];            sort(tmp+1, tmp+4, cmp1);            if(isInterval(tmp[1], tmp[2], tmp[3]))            {                sort(tmp+1, tmp+4, cmp2);                ans[cnt++] = tmp[1].id;                swap(tmp[1], tmp[3]);            }            else                sort(tmp+1, tmp+4, cmp2);        }        sort(ans, ans+cnt);        printf("%d\n", cnt);        for(int i = 0; i < cnt; i++)            printf("%d%c", ans[i], i == cnt-1 ? '\n' : ' ');    }    return 0;}