Hdu 6047 Maximum Sequence【贪心+优先队列】

序列问题求解

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本文介绍了一个关于序列的最大值求解问题，给出了问题描述、输入输出样例及解析，并提供了AC代码实现，通过优先队列来解决序列问题。

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 422 Accepted Submission(s): 226

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;

题目大意：

给你长度为N的A数组，同时给你一个长度为N的B数组，让你每次从B数组中取一个数（bk），在A数组中，从bk到当前位子中，找一个最大的数A【i】-i放到当前位子上。

问怎样分配能够使得后N个数和最大。

思路：

因为越到后边数会越小，贡献也会越小，所以我们直接sort一下B数组，然后按序选择即可。

过程我们可以用优先队列维护一下，也可以用线段树暴力维护。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
#define ll __int64
struct node
{
    int val;
    int pos;
    bool friend operator <(node a,node b )
    {
        return a.val<b.val;
    }
}now,nex;
int a[350000];
int b[350000];
const ll mod=1e9+7;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        priority_queue<node>s;
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)scanf("%d",&b[i]);
        sort(b+1,b+1+n);
        for(int i=1;i<=n;i++)
        {
            now.val=a[i]-i;
            now.pos=i;
            s.push(now);
        }
        ll output=0;
        for(int i=1;i<=n;i++)
        {
            int nowpos=i+n;
            while(!s.empty())
            {
                now=s.top();
                if(now.pos<b[i])s.pop();
                else break;
            }
            now=s.top();
            output+=(ll)now.val;
            nex.pos=nowpos;
            nex.val=now.val-(i+n);
            s.push(nex);
        }
        printf("%I64d\n",(output%mod+mod)%mod);
    }
}