結果

問題 No.2693 Sword
ユーザー hibit_at
提出日時 2024-02-08 21:56:51
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
MLE  
実行時間 -
コード長 4,509 bytes
コンパイル時間 1,375 ms
コンパイル使用メモリ 116,640 KB
最終ジャッジ日時 2025-02-19 02:54:10
ジャッジサーバーID
（参考情報） 		judge2 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 5 MLE * 1 -- * 23
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <limits.h>
#include <map>
#include <math.h>
#include <numeric>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>
#include <stack>
#include <complex>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define rep1(i, n) for (int i = 1; i < n + 1; i++)
#define rev(i, n) for (int i = n - 1; i >= 0; i--)
#define all(A) A.begin(), A.end()
#define itr(A, l, r) A.begin() + l, A.begin() + r
#define debug(var) cout << #var << " = " << var << endl;
typedef long long ll;
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p)
{
    os << "(" << p.first << "," << p.second << ")";
    return os;
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p)
{
    is >> p.first >> p.second;
    return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i] << (i + 1 != (int)v.size() ? " " : "");
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<vector<T>> &v)
{
    for (int i = 0; i < (int)v.size(); i++)
    {
        os << v[i] << endl;
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<vector<vector<T>>> &v)
{
    int n = v.size();
    int m = v[0].size();
    int p = v[0][0].size();
    rep(k, p)
    {
        os << "k = " << k << endl;
        rep(i, n)
        {
            rep(j, m)
            {
                os << v[i][j][k];
                if (j < m - 1)
                {
                    os << " ";
                }
                else
                {
                    os << endl;
                }
            }
        }
    }
    return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v)
{
    for (T &in : v)
        is >> in;
    return is;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, map<T, S> &mp)
{
    for (auto &[key, val] : mp)
    {
        os << key << ":" << val << " ";
    }
    cout << endl;
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, set<T> st)
{
    auto itr = st.begin();
    for (int i = 0; i < (int)st.size(); i++)
    {
        os << *itr << (i + 1 != (int)st.size() ? " " : "");
        itr++;
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, multiset<T> st)
{
    auto itr = st.begin();
    for (int i = 0; i < (int)st.size(); i++)
    {
        os << *itr << (i + 1 != (int)st.size() ? " " : "");
        itr++;
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, queue<T> q)
{
    while (q.size())
    {
        os << q.front() << " ";
        q.pop();
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, deque<T> q)
{
    while (q.size())
    {
        os << q.front() << " ";
        q.pop_front();
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, stack<T> st)
{
    while (st.size())
    {
        os << st.top() << " ";
        st.pop();
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, priority_queue<T> pq)
{
    while (pq.size())
    {
        os << pq.top() << " ";
        pq.pop();
    }
    return os;
}
template <typename T>
ostream &operator<<(ostream &os, priority_queue<T, vector<T>, greater<T>> mpq)
{
    while (mpq.size())
    {
        os << mpq.top() << " ";
        mpq.pop();
    }
    return os;
}
int main()
{
    ll n, p, k;
    cin >> n >> p >> k;
    vector<vector<map<ll, ll>>> dp(n + 1, vector<map<ll, ll>>(k + 1));
    dp[0][0][p] = 1;
    ll lim = ll(1e18) + 1;
    rep(i, n)
    {
        ll t, b;
        cin >> t >> b;
        rep(j, k + 1)
        {
            for (auto [key, val] : dp[i][j])
            {
                dp[i + 1][j][key] = 1;
                if (j + 1 <= k)
                {
                    if (t == 1)
                    {
                        dp[i + 1][j + 1][min(lim, key + b)] = 1;
                    }
                    else
                    {
                        dp[i + 1][j + 1][min(lim, key * 2)] = 1;
                    }
                }
            }
        }
    }
    // cout << dp[n] << endl;
    ll ans = dp[n][k].rbegin()->first;
    if (ans > ll(1e18))
    {
        cout << -1 << endl;
    }
    else
    {
        cout << ans << endl;
    }
}
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