제출 #1246648

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1246648		Lemser	Paint By Numbers (IOI16_paint)	C++20	90 / 100	415 ms	589824 KiB

paint

#include "paint.h"
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
#pragma GCC target("popcnt")
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using lld = long double;
using ii = pair<int,int>;
using pll = pair<ll, ll>;
 
using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vlld = vector<lld>;
 
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define gcd(a,b) __gcd(a,b)
#define sz(x) (int)x.size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fls cout.flush()
 
#define fore(i, l, r) for (auto i = l; i < r; i++)
#define fo(i, n) fore (i, 0, n)
#define forex(i, r, l) for (auto i = r-1; i >= l; i--)
#define ffo(i, n) forex (i, n, 0)
 
bool cmin(ll &a, ll b) { if (b < a) { a=b; return 1; } return 0; }
bool cmax(ll &a, ll b) { if (b > a) { a=b; return 1; } return 0; }
 
const ll INF = 1e18;
const int mod = 1e9 + 7, LOG = 20;

bool is_possible (ll n, ll k, string s, vector<int> c) {
    vector<vll> dp(n+3, vll(k+3, 0));
    s = 'w' + s;
    {auto d = c;
        c = vector<int>(1, 0);
        fo (i, k) c.pb(d[i]);}
    dp[0][0] = 1;
    fore (i, 1, n+1) {
        fo (j, k+1) {
            // elegir nada
            dp[i][j] |= (dp[i-1][j] && (s[i] != 'X' ? 1 : 0));
            // elegir substring
            if (c[j] <= i && j > 0) {
                bool ok = 1;
                forex (x, i+1, i-c[j]+1)
                    ok &= (s[x] != '_');
                ok &= (s[i-c[j]] != 'X');
                if (c[j] == i) dp[i][j] |= ((j == 1) && ok);
                else dp[i][j] |= (dp[i-c[j]-1][j-1] && ok);
            }
            // ___XXX
        }
    }
    return dp[n][k];
}

void prec (ll n, ll k, string s, vector<int> &c, vector<vector<vll>> &dppref, vector<vector<vll>> &dpsuf, vll &pf) {
    s = 'w' + s + 'w';
    {auto d = c;
        c = vector<int>(1, 0);
        fo (i, k) c.pb(d[i]);}
    fore (i, 1, n+1) pf[i] = pf[i-1] + (s[i] == '_');
    dppref[0][0][0] = 1;
    dppref[0][0][1] = 1;
    fore (i, 1, n+1) {
        fo (j, k+1) {
            // elegir nada
            dppref[i][j][0] |= ((dppref[i-1][j][0] || dppref[i-1][j][1]) && (s[i] != 'X' ? 1 : 0));
            // elegir substring
            if (c[j] <= i && j > 0) {
                bool ok = (pf[i] - pf[i-c[j]] == 0 ? 1 : 0);
                ok &= (s[i-c[j]] != 'X');
                if (c[j] == i) dppref[i][j][1] |= ((j == 1) && ok);
                else dppref[i][j][1] |= (dppref[i-c[j]][j-1][0] && ok);
            }
        }
    }
    dpsuf[n+1][0][0] = 1;
    dpsuf[n+1][0][1] = 1;
    forex (i, n+1, 1) {
        fo (j, k+1) {
            // elegir nada
            dpsuf[i][j][0] |= ((dpsuf[i+1][j][0] || dpsuf[i+1][j][1]) && (s[i] != 'X' ? 1 : 0));
            // elegir substring
            if (c[k-j+1] <= n - i + 1 && j > 0) {
                bool ok = (pf[i+c[k-j+1]-1] - pf[i-1] == 0 ? 1 : 0);
                ok &= (s[i+c[k-j+1]] != 'X');
                if (c[k-j+1] == n - i + 1) dpsuf[i][j][1] |= ((j == 1) && ok);
                else dpsuf[i][j][1] |= (dpsuf[i+c[k-j+1]][j-1][0] && ok);
            }
        }
    }
}

string solve_puzzle(string s, vector<int> c) {
    ll n = s.size(), k = c.size();
    string ans = s;
    vector<vector<vll>> dppref(n+3, vector<vll>(k+3, vll(2, 0ll))), dpsuf(n+3, vector<vll>(k+3, vll(2, 0ll)));
    vector<vll> func(k+3, vll(n+3, 0ll));
    vll pf(n+3, 0ll);
    prec(n, k, s, c, dppref, dpsuf, pf);
    fore (j, 1, k+1) {
        fore (i, 1, n+1) {
            func[j][i] = (dppref[i][j][1] && dpsuf[i+1][k-j][0]);
            func[j][i] += func[j][i-1];
        }
    }
    fo (i, n) {
        if (s[i] != '.') continue;
        // cout << "Indice " << i << '\n';
        s[i] = 'X';
        // cout << "checar con X\n";
        bool ok = 0;
        // j = el indice del substring que uso que contenga a i
        fore (j, 1, k+1) {
            ok |= (func[j][min(ll(i+c[j]), n)] > func[j][i] ? 1 : 0);
        }
        // cout << ok << '\n';
        if (ok) ans[i] = 'X';
        // cout << "checar con Y\n";
        ok = 0;
        s[i] = '_';
        fo (j, k+1) {
            ok |= (dppref[i+1][j][0] && (dpsuf[i+2][k-j][0] || dpsuf[i+2][k-j][1]));
        }
        if (ok) ans[i] = (ans[i] == 'X' ? '?' : '_');
        s[i] = '.';
    }
    return ans;
}

컴파일 시 표준 에러 (stderr) 메시지

paint.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~
paint_c.h:1:9: warning: #pragma once in main file
    1 | #pragma once
      |         ^~~~

• Subtask 1: 7 / 7
• Subtask 2: 3 / 3
• Subtask 3: 22 / 22
• Subtask 4: 27 / 27
• Subtask 5: 21 / 21
• Subtask 6: 10 / 10
• Subtask 7: 0 / 10