hdu 6047 Maximum Sequence

2017 Multi-University Training Contest - Team 2 链接：传送门

Maximum SequenceTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 932    Accepted Submission(s): 436Problem DescriptionSteph 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.InputThe 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.OutputFor each test case, print the answer on one line: max{∑2nn+1ai} modulo 109+7。Sample Input48 11 8 53 1 4 2Sample Output27HintFor 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;

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6047

思路：max∑n+12na[i]和最大，就需要最大的数填入n+1的位置，然后一次减小，这样才能保证和最大（使用两个优先队列）。

#include<bits/stdc++.h>#include<queue>#define mod 1000000007using namespace std;struct node{    int p,xia;    bool operator<(const node &a)const    {        return p<a.p;    }};int main(){    int n;    while(~scanf("%d",&n))    {        priority_queue<int,vector<int>,greater<int> >qb;//最小的数先出来        priority_queue<node>qa;//结构体中p,最大的先出来；        long long sum=0;        int a;        node q;        for(int i=1; i<=n; i++)        {            scanf("%d",&a);            q.xia=i;            q.p=a-i;//预处理，得到差值            qa.push(q);        }        for(int i=0; i<n; i++)        {            scanf("%d",&a);            qb.push(a);        }        int k=n+1;        while(!qb.empty())//b数组中的数，从最小的开始找，这样才能保证a[i]最大        {            int x=qb.top();            qb.pop();            q=qa.top();            while(x>q.xia)//将下标之前的数删除，因为x会越来越大，不会再出现，            {                qa.pop();                q=qa.top();            }            qa.pop();            qa.push(q);//重新加入队列，还有可能出现            sum=(sum+q.p)%mod;            q.p=q.p-k;            q.xia=k;            qa.push(q);//新得到数也要加入队列，            k++;        }        while(!qa.empty())            qa.pop();        printf("%d\n",sum);    }    return 0;}