題意：邀請 n 參加聚會，如果在邀請第 i 個人之前，已經成功邀請了 x 個人，並且 li <= x <= ri，那麽第 i 人才會去，問你怎麽排列使得邀請的人最多。

析：對於所有的人，按照 li 進行排序，對於維護一個優先隊列，隊列內是 ri 小的優先，然後枚舉每個時間點，把 li 等於的當前時間點的都放到隊列中，然後取出最優先那個，這就是要邀請的，當然如果隊列頂上的 ri 小於當前時間點，要把它從隊列中刪除。

代碼如下：

#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;
}

HDU 5360 Hiking (貪心)