zoj 3953 三区间重合（贪心）

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

43 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;}