제출 #367315

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
367315		jainbot27	Paint By Numbers (IOI16_paint)	C++17	100 / 100	1242 ms	44072 KiB

paint

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
using namespace std;

#define f first
#define s second
#define pb push_back
#define ar array
#define all(x) x.begin(), x.end()
#define siz(x) (int)x.size()

#define FOR(x, y, z) for(int x = (y); x < (z); x++)
#define ROF(x, z, y) for(int x = (y-1); x >= (z); x--)
#define F0R(x, z) FOR(x, 0, z)
#define R0F(x, z) ROF(x, 0, z)
#define trav(x, y) for(auto&x:y)

using ll = long long;
using vi = vector<int>;
using vl = vector<long long>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;

template<class T> inline bool ckmin(T&a, T b) {return b < a ? a = b, 1 : 0;}
template<class T> inline bool ckmax(T&a, T b) {return b > a ? a = b, 1 : 0;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const char nl = '\n';
const int mxN = 2e5 + 10;
const int mxK = 101;
const int MOD = 1e9 + 7;
const long long infLL = 1e18;

int n, k;
string s[2], res; vi c[2];
int p[mxN][2];
bool dp[mxN][mxK][2], vis[mxN][mxK][2];
bool white[mxN];
int black[mxN];

inline int sum(int l, int r, int T){
    return p[r][T]-(!l?0:p[l-1][T]);
}

inline bool solve(int N, int K, int T){
    if(N < 0){
        if(K == 0) return 1;
        else return 0;
    }
    return dp[N][K][T];
}

inline bool query(int a, int b, int c, int d){
    return solve(a, b, 0)&&solve(c, d, 1);
}

string solve_puzzle(string S2, vi C2){
    s[0] = S2; c[0] = C2;
    s[1] = s[0], c[1] = c[0];
    reverse(all(s[1])); reverse(all(c[1]));
    n = siz(s[0]), k = siz(c[0]);
    c[0].insert(c[0].begin(), 0);
    c[1].insert(c[1].begin(), 0);
    F0R(i, n){
        p[i][0] = (!i?0:p[i-1][0])+(s[0][i]=='_');
        p[i][1] = (!i?0:p[i-1][1])+(s[1][i]=='_');
    }
    F0R(T, 2){
        FOR(K, 0, k+1){
            F0R(i, n){
                if(K == 0){
                    if(s[T][i]!='X') dp[i][K][T] |= solve(i-1, K, T);
                    continue;
                }
                if(s[T][i]!='X')dp[i][K][T]|=solve(i-1, K, T);
                if(i-c[T][K]+1>=0&&sum(i-c[T][K]+1, i, T)==0&&(i-c[T][K]<0||s[T][i-c[T][K]]!='X')) dp[i][K][T] |= solve(i-1-c[T][K], K-1, T);
            }
        }
    }
    F0R(i, n){
        if(s[0][i]!='X'){
            F0R(j, k+1){
                white[i] |= query(i-1, j, (n-1)-(i+1), (k)-j);
            }
        }
    }
    FOR(i, 1, 1+k){
        F0R(j, n+1-c[0][i]){
            if(sum(j, j+c[0][i]-1, 0)==0){
                if((j-1<0||s[0][j-1]!='X')&&(j+c[0][i]>=n||s[0][j+c[0][i]]!='X')){
                    if(query(j-2, i-1, (n-1)-(j+c[0][i]+1), (k)-(i+1)+1)){
                        black[j]++;
                        black[j+c[0][i]]--;
                    }
                }
            }
        }
    }
    res.resize(n);
    F0R(i, n){
        if(black[i]>0&&white[i]) res[i] = '?';
        else if(white[i]) res[i] = '_';
        else {
            res[i] = 'X';
        }
        black[i+1]+=black[i];
    }
    return res;
}

• 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: 10 / 10