HDU 6047 Maximum Sequence (贪心)

Description

Steph is extremely obsessed with “sequence problems” that are usually seen on magazines: Given the sequence 11, 23, 30, 35, what is the next number? Steph always finds them too easy for such a genius like himself until one day Klay comes up with a problem and ask him about it.

Given two integer sequences {ai} and {bi} with the same length n, you are to find the next n numbers of {ai}: an+1…a2n. Just like always, there are some restrictions on $a_{n+1}…a_{2n}$ : for each number ai, you must choose a number bk from {bi}, and it must satisfy $a_i≤\max(a_j-j)~(b_k≤j<i)$ , and any bk can’t be chosen more than once. Apparently, there are a great many possibilities, so you are required to find $\max(\sum^{2n}_{n+1}a_i)$ modulo $10^9+7$ .

Now Steph finds it too hard to solve the problem, please help him.

 

Input

The input contains no more than 20 test cases.

For each test case, the first line consists of one integer n. The next line consists of n integers representing {ai}. And the third line consists of n integers representing {bi}.

$1≤n≤250000, n≤a_i≤1500000, 1≤b_i≤n$

 

Output

For each test case, print the answer on one line: $\max(\sum^{2n}_{n+1}a_i)$ modulo $10^9+7$ 。

 

Sample Input

4
8 11 8 5
3 1 4 2

 

Sample Output

27

 

题意

给出数列 $A$ 和 $B$ ,我们可以从 $B$ 数列中取出一个编号来查找 $A$ 数列中该编号及以后的位置中 $A_i-i$ 的最大值并将其加入末尾,求 $A_{n+1}..A_{2n}$ 的和。

 

思路

题目很简单,但是题意好难懂。(可能是英语差的原因)

贪心思想,既然这样我们每次取 $B$ 中当前编号最小的数字即可,因为这样才有机会让较大的数被加入到较前的位置,然后 $A_i-i$ 作差后的值仍然较大。

具体可以用两个优先队列来实现。

 

AC 代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include<queue>
#include<iostream>
using namespace std ;

typedef __int64 LL;
typedef pair<LL,int> P;

const int mod = 1e9+7;
int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        priority_queue<P>sk;
        priority_queue<int,vector<int>,greater<int> >b;
        LL ans=0;
        for(int i=1; i<=n; i++)
        {
            LL x;
            cin>>x;
            sk.push(P(x-i,i));
        }
        for(int i=1; i<=n; i++)
        {
            int x;
            cin>>x;
            b.push(x);
        }
        for(int ti=n+1; !b.empty(); ti++)
        {
            int i=b.top();
            b.pop();
            while(sk.top().second<i)
                sk.pop();
            P p=sk.top();
            sk.push(P(p.first-ti,ti));
            ans=(ans+p.first)%mod;
        }
        cout<<ans<<endl;
    }
    return 0;
}