Codeforces Round #488 by NEAR (Div. 2) B. Knights of a Polygonal Table
阿新 • • 發佈:2019-01-03
k最大是10, 按照power排序後 維護每個位置的前k大,注意k為0的情況
類似字首和,每個位置的優先佇列按照從小到大的順序排列,同時保證佇列的大小不超過k
#include <bits/stdc++.h>
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; i<n; i++)
{
cin >>node[i].power;
node[i].id=i;
}
for (int i=0; i<n; i++)
{
cin>>node[i].money;
}
sort(node, node+n);
for (int i=0; i<n; i++)
{
if (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; i<n; i++)
{
ll t = node[i].money;
while (!mp[i].empty())
{
t+=mp[i].top();
mp[i].pop();
}
index[node[i].id] = t;
}
for (int i=0; i<n; i++)
{
printf("%lld ", index[i]);
}
return 0;
}