1. 程式人生 > >Codeforces Round #488 by NEAR (Div. 2) B. Knights of a Polygonal Table

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

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; }