제출 #1067024

#제출 시각아이디문제언어결과실행 시간메모리
1067024GrindMachineChorus (JOI23_chorus)C++17
16 / 100
476 ms1048576 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

read the edi some time ago, remember some ideas from there

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    int n,k; cin >> n >> k;
    string s; cin >> s;
    s = "$" + s;
    n <<= 1;

    vector<int> p(n+5);
    rep1(i,n) p[i] = p[i-1]+(s[i]=='A');

    int dp[n+5][n+5][n+5][n+5][n+5];
    memset(dp,0x3f,sizeof dp);
    dp[0][0][0][0][0] = 0;

    rep(i,n){
        int x = p[i+1];
        rep(ignore_cnt,n+1){
            rep(ones,n+1){
                rep(active_ones,n+1){
                    rep(c,n+1){
                        int curr = dp[i][ignore_cnt][ones][active_ones][c];

                        // A
                        amin(dp[i+1][ignore_cnt][ones+1][active_ones+1][c+(active_ones==0)],curr+abs(ones+1-x));

                        // B
                        if(ignore_cnt){
                            amin(dp[i+1][ignore_cnt-1][ones][active_ones][c],curr+abs(ones-x));
                        }
                        else{
                            if(active_ones){
                                amin(dp[i+1][active_ones-1][ones][0][c],curr+abs(ones-x));
                            }
                        }
                    }
                }
            }
        }
    }

    int ans = inf1;
    rep(c,k+1){
        amin(ans,dp[n][0][n/2][0][c]);
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}
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