제출 #410053

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
410053		534351	카니발 티켓 (IOI20_tickets)	C++17	39 / 100	3090 ms	119852 KiB

tickets

#include "tickets.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int) (x).size())

const int MAXN = 1513;
const long long LLINF = 3e18;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;

int N, M, K;
ll grid[MAXN][MAXN];
ll smallest[MAXN][MAXN], biggest[MAXN][MAXN];
vl dp[MAXN];//at row i, we have j -s. what's the best we could get?
vi parent[MAXN];
int mm[MAXN], pp[MAXN];
int mtake[MAXN], ptake[MAXN];
bitset<MAXN> rec;
vector<vi> ans;

long long find_maximum(int k, vector<vi> xx)
{
    K = k; N = SZ(xx); M = SZ(xx[0]);
    ans.resize(N);
    FOR(i, 0, N)
    {
        ans[i].resize(M);
        FOR(j, 0, M)
        {
            ans[i][j] = -1;
            grid[i][j] = xx[i][j];
        }
        FOR(j, 0, M)
        {
            smallest[i][j + 1] = smallest[i][j] + grid[i][j];
            biggest[i][j + 1] = biggest[i][j] + grid[i][M - 1 - j];
        }
    }
    dp[0].PB(0);
    FOR(i, 0, N)
    {
        //you're choosing K numbers.
        dp[i + 1].resize(K * (i + 1) + 1);
        parent[i + 1].resize(K * (i + 1) + 1);
        fill(ALL(dp[i + 1]), -LLINF);
        FOR(j, 0, K * i + 1)
        {
            FOR(k, 0, K + 1) //choose k -s
            {
                ll cand = dp[i][j] - smallest[i][k] + biggest[i][K - k];
                if (cand > dp[i + 1][j + k])
                {
                    dp[i + 1][j + k] = cand;
                    parent[i + 1][j + k] = k;
                    // cerr << "PARENT " << i + 1 << ' ' << j + k << " = " << parent[i + 1][j + k] << endl;
                }
            }
        }
    }
    k = K * (N / 2);
    FORD(i, N + 1, 1)
    {
        // cerr << "parent " << i << ' ' << k << " = " << parent[i][k] << endl;
        mtake[i - 1] = parent[i][k];
        k -= parent[i][k];
    }
    FOR(i, 0, N)
    {
        ptake[i] = K - mtake[i];
        mm[i] = 0;
        pp[i] = M - 1;
        // cerr << pp[i] << ' ' << M - 1 - ptake[i] << endl;
        // cerr << "ptake " << i << " = " << ptake[i] << endl;
    }
    FOR(i, 0, K)
    {
        // cerr << "work " << i << endl;
        int minuses = 0, pluses = 0;
        FOR(j, 0, N)
        {
            rec[j] = false;
            if (mm[j] == mtake[j])
            {
                //you have to make it a plus
                // cerr << "make " << j << " plus\n";
                pluses++;
                ans[j][pp[j]] = i;
                pp[j]--;
                rec[j] = true;
            }
            else if (pp[j] == M - 1 - ptake[j])
            {
                // cerr << "make " << j << " mnus\n";
                minuses++;
                ans[j][mm[j]] = i;
                mm[j]++;
                rec[j] = true;
            }
        }
        FOR(j, 0, N)
        {
            if (rec[j]) continue;
            if (pluses < N / 2)
            {
                // cerr << "make " << j << " PLus\n";
                pluses++;
                ans[j][pp[j]] = i;
                pp[j]--;
                rec[j] = true;
            }
            else
            {
                // cerr << "make " << j << " MNus\n";
                minuses++;
                ans[j][mm[j]] = i;
                mm[j]++;
                rec[j] = true;
            }
        }
        //for each row that already got all -s and all +s factor it in.
    }
    //as long as we make sure to choose enough +s and enough -s we shouldn't need to worry right.
    //keep a set of (value, position.)
    //i think you always want to take x[0] or x[end] right?
    //so if x is 1, then
	allocate_tickets(ans);
	return dp[N][K * (N / 2)];
}

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• Subtask 2: 16 / 16
• Subtask 3: 0 / 14
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• Subtask 5: 12 / 12
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• Subtask 7: 0 / 10