HDU 5360 Hiking

Problem Description

There are  n  soda conveniently labeled by  1,2,…,n . beta, their best friends, wants to invite some soda to go hiking. The  i -th soda will go hiking if the total number of soda that go hiking except him is no less than  li  and no larger than  ri . beta will follow the rules below to invite soda one by one:
1. he selects a soda not invited before;
2. he tells soda the number of soda who agree to go hiking by now;
3. soda will agree or disagree according to the number he hears.

Note: beta will always tell the truth and soda will agree if and only if the number he hears is no less than  li  and no larger than  ri , otherwise he will disagree. Once soda agrees to go hiking he will not regret even if the final total number fails to meet some soda's will.

Help beta design an invitation order that the number of soda who agree to go hiking is maximum.

 

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 contains an integer  n   (1≤n≤105) , the number of soda. The second line constains  n  integers  l1,l2,…,ln . The third line constains  n  integers  r1,r2,…,rn .  (0≤li≤ri≤n)
It is guaranteed that the total number of soda in the input doesn't exceed 1000000. The number of test cases in the input doesn't exceed 600.

 

Output

For each test case, output the maximum number of soda. Then in the second line output a permutation of  1,2,…,n  denoting the invitation order. If there are multiple solutions, print any of them.

 

Sample Input

   
   
   
   

    
    
    
    4
8
4 1 3 2 2 1 0 3
5 3 6 4 2 1 7 6
8
3 3 2 0 5 0 3 6
4 5 2 7 7 6 7 6
8
2 2 3 3 3 0 0 2
7 4 3 6 3 2 2 5
8
5 6 5 3 3 1 2 4
6 7 7 6 5 4 3 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7
1 7 6 5 2 4 3 8
8
4 6 3 1 2 5 8 7
7
3 6 7 1 5 2 8 4
0
1 2 3 4 5 6 7 8
   
   
   
   

优先级队列，按右端排序，然后剩下的随便搞。
#include<cstdio>
#include<cmath>
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const ll maxn=100005;
int T,n,m,l[maxn],r[maxn],ans[maxn],tot,f[maxn];

struct lr
{
    int l,r,id;
    lr(int l,int r,int id):l(l),r(r),id(id){};
    bool operator <(const lr& a)const
    {
        if (r==a.r) return l>a.l;
        return r>a.r;
    }
};

vector<lr> p[maxn];

int main()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);    tot=0;
        for (int i=1;i<=n;i++) scanf("%d",&l[i]);
        for (int i=1;i<=n;i++) scanf("%d",&r[i]);
        for (int i=0;i<=n;i++) p[i].clear();
        for (int i=1;i<=n;i++) p[l[i]].push_back(lr(l[i],r[i],i));
        priority_queue<lr> que;
        for (int i=0;i<=n;)
        {
            for (int j=0;j<p[i].size();j++) que.push(lr(p[i][j]));
            while (!que.empty()&&que.top().r<i) que.pop();
            if (!que.empty()) {ans[tot++]=que.top().id; que.pop(); i++;}
            else break;
        }
        printf("%d\n",tot);
        if (tot) 
        {
            memset(f,0,sizeof(f));
            printf("%d",ans[0]);
            for (int i=1;i<tot;i++) printf(" %d",ans[i]);
            for (int i=0;i<tot;i++) f[ans[i]]=1;
            for (int i=1;i<=n;i++) if (!f[i]) printf(" %d",i);
        }
        else for (int i=1;i<=n;i++) 
        {
            if (i>1) printf(" ");
            printf("%d",i);
        }
        printf("\n");
    }
    return 0;
}