hdu 6047 Maximum Sequence (贪心)

Maximum Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2325    Accepted Submission(s): 1091

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;

分析：贪心思想，将b数组升序排序后，此时第一次选的bi最小，在前n个a数组里能选到的数就最大，a数组往后加的第一个位置放的数肯定是最大的，所以，虽然每次从a中找数时，随着bi增大，坐标越来越靠后，但是最大的数已经筛选出来了，放在a[n+1]了，这符合贪心思想。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int MOD=1e9+7;

int a[250005],b[250005];
int mm[250005];

int main()
{
    int n;
    long long ans;
    while(~scanf("%d",&n))
    {
        ans=0;
        memset(mm,0,sizeof(mm));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]-=i;
        }
        for(int i=1;i<=n;i++)
            scanf("%d",&b[i]);
        sort(b,b+n);
        for(int i=n;i>0;i--)
            mm[i]=max(mm[i+1],a[i]);
        long long mx=(long long)mm[b[1]];
        ans=(ans+mx)%MOD;
        mx-=n+1;
        for(int i=2;i<=n;i++)
        {
            long long tem=(long long)mm[b[i]];
            tem=max(tem,mx);
            ans=(ans+tem)%MOD;
        }
        printf("%lld\n",ans);
    }
    return 0;
}