hdu6047 Maximum Sequence 2017多校二1003

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1077    Accepted Submission(s): 503

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
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define maxn 260000#define Mod 1000000007#define ll long longusing namespace std;struct data{    ll val,poi;};bool operator<(data a,data b){    return a.val<=b.val;}ll b[maxn];int main(){    ll n;    while(~scanf("%lld",&n))    {        priority_queue<data,vector<data> > que;        while(!que.empty())            que.pop();        for (ll k=1;k<=n;k++)        {            data save;            scanf("%lld",&save.val);            save.val-=k;save.poi=k;            que.push(save);        }        for (ll k=1;k<=n;k++)            scanf("%lld",&b[k]);        sort(b+1,b+n+1);        ll sum(0);        for (ll k=1;k<=n;k++)        {            while(que.top().poi<b[k])                que.pop();            sum=(sum+que.top().val)%Mod;            data ss;            ss.val=que.top().val;            ss.poi=n+k;            ss.val=ss.val-ss.poi;            que.push(ss);        }        printf("%lld\n",sum);    }    return 0;}