CodeForces-1315D Recommendations

传送门：CodeForces-1315D

题目描述

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 $a_i$ 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 $t_i$ 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.

输入描述

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 $a_i$ — the number of publications of $i$-th category selected by the batch algorithm $(1≤a_i≤10^9)$.

The third line of input consists of n integers $t_i$ — time it takes for targeted algorithm to find one new publication of category $i$ $(1≤t_i≤10^5)$.

输出描述

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

思路分析

贪心的想，如果有很多个相同的元素，那么将花费最大的留在原地，其他+1肯定是最优的，可以用set来维护最值

样例输入

5
3 7 9 7 8
5 2 5 7 5

样例输出

6

AC代码

#include <functional>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <vector>
#include <string>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define vi vector<int>
#define lowbit(x) x&(-x)
#define pii pair<int,int>
#define all(x) x.begin(),x.end()
#define mem(a,b) memset(a,b,sizeof(a))
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0);

using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;

multiset<ll>q;
pair<ll, ll>a[N];

int main() {
    IOS;
#ifdef xiaofan
    freopen("1.in", "r", stdin);
    freopen("1.out", "w", stdout);
#endif

    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
        cin >> a[i].fi;
    for (int i = 0; i < n; i++)
        cin >> a[i].se;
    sort(a, a + n);

    int now = -1;
    ll ans = 0, sum = 0;
    for (int i = 0; i < n; i++) {
        while (q.size() && now < a[i].fi) {
            ll ma = *q.rbegin();
            ans += sum - ma;
            sum -= ma;
            q.erase(q.find(ma));
            now++;
        }
        q.insert(a[i].se);
        sum += a[i].se;
        now = a[i].fi;
    }

    while (q.size()) {
        ll ma = *q.rbegin();
        ans += sum - ma;
        sum -= ma;
        q.erase(q.find(ma));
    }

    cout << ans << endl;






    return 0;
}
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