HDU 6047 Maximum Sequence

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1152    Accepted Submission(s): 537

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

/*
* @Author: Lyucheng
* @Date:   2017-07-28 15:53:31
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-07-28 17:38:20
*/
/*
 题意：给你序列a,b，长度为n,让你构造a序列n+1~n*2的元素，有一个规则：
    ai≤max{aj-j│bk≤j<i}

 思路：线段树维护a的最大值
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

#define MAXN 250009
#define lson i*2,l,m
#define rson i*2+1,m+1,r
#define INF 0x3f3f3f3f
#define LL long long
const LL MOD = 1e9+;

using namespace std;

int n;
int a[MAXN];
int b[MAXN];
int sum[MAXN*];

void pushup(int i,int l,int r){
    sum[i]=max(sum[i*],sum[i*+]);
}

void build(int i,int l,int r){
    if(l==r){
        if(l<=n)
            sum[i]=a[l]-l;
        return;
    }
    int m=(l+r)/;
    build(lson);
    build(rson);
    pushup(i,l,r);
}

void update(int key,int val,int i,int l,int r){
    if(l==r){
        sum[i]=val;
        return ;
    }
    int m=(l+r)/;
    if(m>=key) update(key,val,lson);
    else update(key,val,rson);
    pushup(i,l,r);
}

int query(int ql,int qr,int i,int l,int r){
    if(ql<=l&&r<=qr){
        return sum[i];
    }
    int m=(l+r)/;
    int res=-;
    if(m>=ql) res=max(res,query(ql,qr,lson));
    if(m<qr) res=max(res,query(ql,qr,rson));
    return res;
}

int main(){
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    while(scanf("%d",&n)!=EOF){
        for(int i=;i<=n;i++){
            scanf("%d",&a[i]);
        }
        for(int i=;i<=n;i++){
            scanf("%d",&b[i]);
        }
        build(,,n*);
        sort(b+,b+n+);
        LL res=;
        for(int i=n+;i<=*n;i++){
            int l=b[i-n];//b中剩余最小的
            int cur=query(l,i-,,,n*);//a中最大的
            update(i,cur-i,,,n*);
            res+=cur;
            res%=MOD;
        }
        printf("%lld\n",res);
    }
    return ;
}