题意：邀请 n 参加聚会，如果在邀请第 i 个人之前，已经成功邀请了 x 个人，并且 li <= x <= ri，那么第 i 人才会去，问你怎么排列使得邀请的人最多。
析：对于所有的人，按照 li 进行排序，对于维护一个优先队列，队列内是 ri 小的优先，然后枚举每个时间点，把 li 等于的当前时间点的都放到队列中，然后取出最优先那个，这就是要邀请的，当然如果队列顶上的 ri 小于当前时间点，要把它从队列中删除。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define be begin()
#define ed end()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,n,x) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.in", "r", stdin)
#define freopenw freopen("out.out", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 20;
const int maxm = 1e6 + 10;
const LL mod = 1000000000000000LL;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
inline int readInt(){ int x; scanf("%d", &x); return x; } struct Node{
int l, r, id;
bool operator < (const Node &rhs) const{
return r > rhs.r;
}
}; Node a[maxn];
bool vis[maxn]; int main(){
int T; cin >> T;
while(T--){
scanf("%d", &n);
for(int i = 0; i < n; ++i){
scanf("%d", &a[i].l);
a[i].id = i;
}
for(int i = 0; i < n; ++i) scanf("%d", &a[i].r);
sort(a, a + n, [&](Node a, Node b){ return a.l < b.l; });
priority_queue<Node> pq;
vector<int> ans;
int k = 0;
for(int i = 0; i < n; ++i){
while(ans.sz == i && k < n && a[k].l == i) pq.push(a[k++]);
while(!pq.empty() && pq.top().r < i) pq.pop();
if(pq.empty()) continue;
ans.pb(pq.top().id); pq.pop();
}
printf("%d\n", ans.sz);
int cnt = 0; ms(vis, 0);
for(int i = 0; i < ans.sz; ++i){
vis[ans[i]] = 1;
if(cnt) printf(" %d", ans[i] + 1);
else printf("%d", ans[i] + 1);
cnt = 1;
}
for(int i = 0; i < n; ++i) if(!vis[i]){
if(cnt) printf(" %d", i + 1);
else printf("%d", i + 1);
cnt = 1;
}
printf("\n");
}
return 0;
}