HDU5360——优先队列——Hiking

/*
大意：n个人参加一个活动的条件有人数限制，l和r，问你最多有多少个人能一起去，把不能参加的人随便输出
用优先队列维护满足的情况
把块从0分到n，贪心地把r最小的并且满足大于等于l的值取出，注意判断如果这一次没有了，下一次也肯定没有这种情况
*/
/************************************************
Author        :powatr
Created Time  :2015-8-6 14:45:47
File Name     :b.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAX = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct edge{
    int l, r, id;
}a[MAX];
struct edge1{
    int x, y;
};
edge1 P;

struct cmp1
{
    bool operator () (edge1 x, edge1 y)
    {
        return x.x > y.x;
    }
};

int b[MAX];
int vis[MAX];
int n;
vector<edge1>G[MAX];

int main(){
    int T;
    priority_queue < edge1 , vector<edge1> , cmp1 > q;
    scanf("%d", &T);
    while(T--){
        memset(vis, 0, sizeof(vis));
        scanf("%d", &n);
        memset(a, 0, sizeof(a));
        for(int i = 1; i <= n; i++){
            scanf("%d",&a[i].l);
        }
        for(int i = 0 ; i <= 100000; i++)
            G[i].clear();
        for(int i = 1; i <= n; i++)
            scanf("%d",&a[i].r);
        for(int i = 1; i <= n; i++)
            a[i].id = i;
        // sort(a + 1, a + n + 1);

         while(!q.empty()) q.pop();

        for(int i = 1; i <= n; i++)
            G[a[i].l].push_back((edge1){a[i].r, a[i].id});

        for(int i = 0 ; i < G[0].size(); i++)
            q.push((edge1){G[0][i].x,G[0][i].y});
        int ans = 1;
        edge1 m;
        if(!q.empty()){
            m = q.top();
         if(m.x >= 0){
           b[ans++] = m.y;
            q.pop();
         }
        }
            
        for(int i = 1; i <= n; i++){

            for(int j = 0; j < G[i].size(); j++)
                q.push((edge1){G[i][j].x, G[i][j].y});
           // if(ans < i + 2) break;
            while(!q.empty()){
                m = q.top();
               if(ans < i + 1) break;
               if(ans == i + 2) break;
                if(m.x >= i){
                q.pop();
                b[ans++] = m.y;
            }
                else q.pop();
            }
        }
        printf("%d\n", ans - 1);
        for(int i = 1; i < ans ; i++){
            printf("%d ", b[i]);
            vis[b[i]] = 1;
        }
         for(int i = 1 ; i <= n; i++)
             if(!vis[i])
                    printf("%d ", i);
         puts("");
    }
    return 0;
}