hdu6047Maximum Sequence（优先队列）

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1829    Accepted Submission(s): 857

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
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;
题意：给定相同的长度串a[i]，b[i]，在b中取数，每次不能相同，使得满足ai≤max{aj-j│bk≤j<i}，例如a  8 11 8 5b  3 1 4 2b取2，则让a的第二个位置到a的第五个位置-2进行第五个位置的值选取，选取值为a[i]-i，其中i为当前位置。这样取1时，a_2…a_4中取一个值给a[5]}取出来的值，要求和最大，取余1e+7.
#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <queue>#include <iostream>#define LL long long#define mod 1000000000+7using namespace std;struct node{    LL dis,ip,num;};struct cmp{    bool operator()(node a,node b)    {        return a.dis<b.dis;    }};LL b[250010];int main(){    LL n;    node a;    while(scanf("%lld",&n)!=EOF)    {        priority_queue<node,vector<node>,cmp>Q;        for(int i=1; i<=n; i++)        {            scanf("%d",&a.num);            a.ip=i;            a.dis=a.num-a.ip;            Q.push(a);            //c[i]=Q.top();        }        for(int i=1; i<=n; i++)            scanf("%lld",&b[i]);        sort(b+1,b+1+n);        long long res=0;        for(int i=1; i<=n; i++)        {            while(Q.top().ip<b[i])                Q.pop();            node t=Q.top();            res+=t.dis;            res%=mod;            t.ip=n+i;            t.num=t.dis;            t.dis-=t.ip;            Q.push(t);        }        printf("%lld\n",res);    }    return 0;}