Maximum Sequence（hdu 6047）

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 567    Accepted Submission(s): 297

Problem 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  an+1…a2n : for each number  ai , you must choose a number  bk  from {bi}, and it must satisfy  ai ≤max{ aj -j│ bk ≤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{ ∑2nn+1ai } modulo  109 +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{ ∑2nn+1ai } modulo  109 +7。

 

Sample Input

  

   4
8 11 8 5
3 1 4 2
  

 

Sample Output

  

   27

   

    

     Hint
    
For the first sample:
1. Choose 2 from {bi}, then a_2…a_4 are available for a_5, and you can let a_5=a_2-2=9; 
2. Choose 1 from {bi}, then a_1…a_5 are available for a_6, and you can let a_6=a_2-2=9;

   
    
  

 

Source

2017 Multi-University Training Contest - Team 2

 

// 题意 ：先输入一个n表示a[]和b[]的元素个数，然后输入a[]，b[]，每次从b[]里挑1个（每个b里的元素只能用1次），现在要算an->a2n，且满足 a i ≤max{ aj -j│ bk ≤j<i}，计算an->a2n的和的最大值模上10^9+7。

//思路：先计算出a里每个元素的价值ai - i，放入优先队列，大的在前，再把b按从大到小排序，因为b越小，能选到的值就越大，让an->a2n 越前的元素选的b值越小，这样a后面的值能选择的值也就大一点。

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

const int MAX = 250000 + 100;
const int mod = 1e9 + 7;
const int INF = -1e9;

typedef struct {
	int val;
	int id;
}Point;

int n;
int a[MAX], b[MAX];
Point point[MAX];
priority_queue<Point>q;

bool operator < (const Point &t1, const Point &t2) {
	return t1.val < t2.val;
 }

int main()
{
	while (scanf("%d", &n) != EOF)
	{
		long long sum = 0;
		int ans = 0;
		while (!q.empty())
			q.pop();
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
			point[i].val = a[i] - i;
			point[i].id = i;
			q.push(point[i]);
		}
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &b[i]);
		}
		sort(b + 1, b + n + 1);
		for (int i = 1; i <= n; i++)
		{
			Point now = q.top();
			if (b[i] <= now.id)
			{
				sum += now.val;
				sum = sum%mod;
				Point p;
				p.val = now.val - n - i;
				p.id = n + i;
				q.push(p);
			}
			else
			{
				q.pop();
				while (!q.empty())
				{
					Point x = q.top();
					if (b[i] <= x.id)
					{
						sum += x.val;
						sum = sum%mod;
						Point p;
						p.val = x.val - n - i;
						p.id = n + i;
						q.push(p);
						break;
					}
					else
						q.pop();
				}
			}
		}
		sum = sum%mod;
		printf("%lld\n", sum);
	}
	return 0;
}

10

109 +7