Codeforces Round #488 by NEAR (Div. 2) B. Knights of a Polygonal Table

k最大是10， 按照power排序后 维护每个位置的前k大，注意k为0的情况
类似前缀和，每个位置的优先队列按照从小到大的顺序排列，同时保证队列的大小不超过k

#include 
using namespace std;

const int maxn = 1e6 + 10;
const int N = maxn;
typedef long long ll;
#define gcd __gcd


struct Node
{
    int power;
    int id;
    int money;
    bool operator<(const Node& rhs)
    {
        return power < rhs.power;
    }
} node[N];

ll index[N];

map<int, priority_queue<int, vector<int>, greater<int> > > mp;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    for (int i=0; icin>>node[i].power;
        node[i].id=i;
    }

    for (int i=0; icin>>node[i].money;
    }

    sort(node, node+n);

    for (int i=0; iif (i==0);
        else
        {
            mp[i] = mp[i-1];
            if (mp[i].size() >= k)
            {
                if (k)
                {
                    int tmp = mp[i].top();
                    if (tmp < node[i-1].money)
                    {
                        mp[i].pop();
                        mp[i].push(node[i-1].money);
                    }
                }

            }
            else
            {
                mp[i].push(node[i-1].money);
            }
        }
    }


    for (int i=0; iwhile (!mp[i].empty())
        {
            t+=mp[i].top();
            mp[i].pop();
        }

        index[node[i].id] = t;
    }

    for (int i=0; iprintf("%lld ", index[i]);
    }



    return 0;
}