HDU 6047 Maximum Sequence

                                 Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1887    Accepted Submission(s): 671

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

48 11 8 53 1 4 2

 

Sample Output
27

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;

 

题意：

给你2个队列，一个队列a1到an，另一个b1-bn （1≤b_i≤n）。

让你再在第一个队列中添加n个数，添加方式：选一个bi，从a[bi]到a[end]（已添加的也算，但是仔细分析其实没有用）,选max（a[i]-i）添加。

求n个数和最大。注意：bi用过了就不能用了。

题目分析：预处理ai=ai-i。max数组用来保存第i到第n里面的最大值，可以倒着算出。具体做法：

可以对b[]数组从小到大排序，每次使用最前面的bi，可以保证a[n+1]是最大的，而且分析可得a[n+1]到a[2n]是不严格单调递减。

所以每次添加数，只要用max[bi]和a[n+1]-(n+1)比较，算出最大值，就是要添加数了，这样就可以算出答案。

#include<bits/stdc++.h>using namespace std;const int mod=1e9+7;const int N=250000+10;int a[N];int b[N];int Max[N];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        long long ans=0;        for(int i=1;i<=n;i++)///预处理a[i]-i，a[i]存储的就是差值        {            scanf("%d",&a[i]);            a[i]-=i;        }        for(int i=1; i<=n; i++)        {            scanf("%d",&b[i]);        }        sort(b+1,b+1+n);///对b[]进行从小到大的排序        memset(Max,0,sizeof Max);        Max[n]=a[n];        for(int i=n-1; i>=1; i--)///先用Max[]保存a[i]-i        {                        ///遍历后呈现出不严格单调递减            if(a[i]>Max[i+1])            {                Max[i]=a[i];            }            else                Max[i]=Max[i+1];        }                int an1;        int mx=Max[b[1]];        an1=mx-(n+1);        (ans+=(long long)mx)%=mod;        for(int i=2;i<=n; i++)        {            mx=Max[b[i]];            mx=max(mx,an1);            (ans+=(long long)mx)%=mod;        }        printf("%lld\n",ans);    }}