Codeforces Round 623(Div. 2,based on VK Cup 2019-2020 - Elimination Round,Engine)D. Recommendations

VK news recommendation system daily selects interesting publications of one of n disjoint categories for each user. Each publication belongs to exactly one category. For each category i batch algorithm selects ai publications.

The latest A/B test suggests that users are reading recommended publications more actively if each category has a different number of publications within daily recommendations. The targeted algorithm can find a single interesting publication of i-th category within ti seconds.

What is the minimum total time necessary to add publications to the result of batch algorithm execution, so all categories have a different number of publications? You can’t remove publications recommended by the batch algorithm.

Input

The first line of input consists of single integer n — the number of news categories (1≤n≤200000).

The second line of input consists of n integers ai — the number of publications of i-th category selected by the batch algorithm (1≤ai≤109).

The third line of input consists of n integers ti — time it takes for targeted algorithm to find one new publication of category i (1≤ti≤105).

Output

Print one integer — the minimal required time for the targeted algorithm to get rid of categories with the same size.

Examples

inputCopy

5

3 7 9 7 8

5 2 5 7 5

outputCopy

6

inputCopy

5

1 2 3 4 5

1 1 1 1 1

outputCopy

0

Note

In the first example, it is possible to find three publications of the second type, which will take 6 seconds.

In the second example, all news categories contain a different number of publications.

优先队列，每次都让某一位上的全职最小的加1，2，3然后处理。

#include <bits/stdc++.h>
using namespace std;
int n, t, a[3000000];

struct node
{
    int first, second;
    bool operator<(const node &b) const
    {
        if (first == b.first)
            return second < b.second;
        else
            return first > b.first;
    }
};
priority_queue<node> ms;
int main()
{
    cin >> n;
    for (int i = 0; i < n; ++i)
    {
        cin >> a[i];
    }
    for (int i = 0; i < n; ++i)
    {
        int b;
        cin >> b;
        ms.push(node{a[i], b});
    }
    long long cnt = 0;
    while (ms.size())
    {
        auto po = ms.top();
        ms.pop();
       //cout << po.first << " " << po.second << endl;
        //cout << 1 << endl;
        if (ms.empty())
            break;
            int f=1;
        while (po.first == ms.top().first && ms.size())
        {
            auto pi = ms.top();
            //cout<<pi.first + 1<<" "<<pi.second<<endl;
            ms.pop();
            ms.push(node{pi.first + f, pi.second});
            cnt += pi.second;
            f++;
          //  cout << cnt << endl;
            // cout<<ms.top().first<<endl;
            // cout << ms.size() << endl;
        }
    }
    cout << cnt << endl;
}

上分代码的思路确实有问题，时间复杂度太高。思路差不多，都是先处理权值大的，让权值小移动。

#include <bits/stdc++.h>
using namespace std;
int n, t;

struct node
{
    int data;
    long long  cost;
    bool operator<(const node &b) const
    {
        if (cost == b.cost)
            return data < b.data;
        else
            return cost > b.cost;
    }
} a[3000000];
map<int, int> fa;
int find(int a)
{
    if (fa[a] == 0)
        return a;
    else
        return fa[a] = find(fa[a]);
}
void unite(int x, int y)
{
    x = find(x);
    y = find(y);
    if (x != y)
        fa[x] = y;
}
int main()
{
    cin >> n;
    fa.clear();
    for (int i = 0; i < n; ++i)
        scanf("%d", &a[i].data);

    for (int i = 0; i < n; ++i)
        scanf("%lld", &a[i].cost);
    sort(a, a + n);
    long long ans = 0;
    for (int i = 0; i < n; i++)
    {
        int x=find(a[i].data) ;
        ans += a[i].cost * (x- a[i].data);
        unite(x,x+1);
    }
    cout << ans << endl;
}