이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF (int)1e18
#define endl '\n'
const int mod = 1000 * 1000 * 1000 + 7;
const int N = 100005;
#define f first
#define s second
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define rrep(i, a, b) for(int i = (a); i > (b); i--)
#define vi vector<int>
#define pii pair<int, int>
#define all(x) (x).begin(), (x).end()
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());
/*
*/
void Solve() {
int s , n; cin>>s>>n;
vector<vector<pair<int,int>> > a(s+1);
for (int i = 0; i < n; ++i) {
int v , w , k; cin>>v>>w>>k;
if (v>0 and w <= s){
a[w].push_back({v,k});
}
}
vector<pii> item;
for (int i = 1; i <=s; ++i) {
sort(a[i].begin(), a[i].end());
int lim = s/ i;
while (!a[i].empty() and lim > 0){
if (a[i].back().second==0){ //count of elments is zero
a[i].pop_back();
continue;
}
a[i].back().second--;
item.push_back({a[i].back().first,i});
lim--;
}
}
//now its normal knapsack problem, where there s log s elment
int sz = (int)item.size();
vector<vector<int> > dp(sz, vector<int> (s+1));
// debug(item);
for (int i = 0; i < sz; ++i) {
for(int j = item[i].second ; j <= s ; j++){
if (i==0){
dp[i][j] = max(dp[i][j] , item[i].first);
continue;
}
dp[i][j] = max(dp[i-1][j-item[i].second] + item[i].first , dp[i-1][j]);
}
}
// debug(dp);
cout<<dp[sz-1][s]<<endl;
}
int32_t main() {
auto begin = std::chrono::high_resolution_clock::now();
ios_base::sync_with_stdio(false);
cin.tie(0);
int t = 1;
// cin >> t;
for (int i = 1; i <= t; i++) {
//cout << "Case #" << i << ": ";
Solve();
}
auto end = std::chrono::high_resolution_clock::now();
auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\n";
return 0;
}
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