이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "paint.h"
#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
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> vll;
#define FOR(i,j,k,in) for(int i=(j); i < (k);i+=in)
#define FORD(i,j,k,in) for(int i=(j); i >=(k);i-=in)
#define REP(i,b) FOR(i,0,b,1)
#define REPD(i,b) FORD(i,b,0,1)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(x) begin(x), end(x)
#define MANY_TESTS int tcase; cin >> tcase; while(tcase--)
const double EPS = 1e-9;
const int MOD = 1e9+7;
const ll INFF = 1e18;
const int INF = 1e9;
const ld PI = acos((ld)-1);
const vi dy = {1, 0, -1, 0, -1, 1, 1, -1};
const vi dx = {0, 1, 0, -1, -1, 1, -1, 1};
void DBG(){cout << "]" << endl;}
template<typename T, typename ...U> void DBG(const T& head, const U... args){ cout << head << "; "; DBG(args...); }
#define dbg(...) cout << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__);
#define chk() cout << "Check at line(" << __LINE__ << ") hit." << endl;
template<class T, unsigned int U>
ostream& operator<<(ostream& out, const array<T, U> &v){out << "["; REP(i, U) out << v[i] << ", "; out << "]"; return out;}
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "[" << par.first << ";" << par.second << "]"; return out;}
template <class T>
ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; }
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; }
template<class T>
ostream& operator<<(ostream& out, const vector<T> &v){ out << "["; REP(i, v.size()) out << v[i] << ", "; out << "]"; return out;}
template<class T>
istream& operator>>(istream& in, vector<T> &v){ for(auto &x : v) in >> x; return in; }
std::string solve_puzzle(std::string s, std::vector<int> c) {
int k = c.size();
int n = (int)s.length();
auto solve_dp = [&](string ts){
vector<vector<int>> dp(n + 1, vector<int>(k + 1, 0));
vector<int> ps(n + 1);
ps[0] = 0;
FOR(i, 1, n + 1, 1){
ps[i] += ps[i - 1];
if(ts[i - 1] != '_')
ps[i]++;
}
dp[0][0] = 1;
FOR(i, 1, n + 1, 1){
REP(j, k + 1){
if(dp[i - 1][j] && ts[i - 1] != 'X')
dp[i][j] = 1;
if(j == 0) continue;
bool can1 = false;
int lst = i - c[j - 1];
if(lst < 0) continue;
if(lst >= 0 && ps[i] - ps[lst] == c[j - 1])
can1 = true;
bool can2 = false;
if(lst == 0 || ts[lst - 1] != 'X')
can2 = true;
if(can1 && can2){
if(lst == 0){
if(j == 1)
dp[i][j] = 1;
}
else if(dp[lst - 1][j - 1])
dp[i][j] = 1;
}
}
}
return dp;
};
string ans(n, '?');
vector<int> ps(n + 1);
ps[0] = 0;
FOR(i, 1, n + 1, 1){
ps[i] += ps[i - 1];
if(s[i - 1] != '_')
ps[i]++;
}
/*
REP(i, n){
if(s[i] == 'X')
ans[i] = 'X';
else if(s[i] == '_')
ans[i] = '_';
else{
ns[i] = 'X';
auto dp = solve_dp(ns);
bool canx = dp[n][k];
ns[i] = '_';
dp = solve_dp(ns);
bool cany = dp[n][k];
ns[i] = s[i];
if(canx && !cany)
ans[i] = 'X';
else if(cany && !canx)
ans[i] = '_';
}
}
*/
auto dp1 = solve_dp(s);
reverse(all(s)); reverse(all(c));
auto dp2 = solve_dp(s);
reverse(all(dp2));
vector<bool> canx(n, false);
vector<bool> cany(n, false);
reverse(all(c));
reverse(all(s));
REP(i, k){
int pp = 0;
REP(j, n - c[i] + 1){
int right = j + c[i];
bool can3 = false;
if(ps[right] - ps[j] == c[i])
can3 = true;
bool can1 = false;
if(j == 0){
if(i == 0)
can1 = true;
}
else{
if(s[j - 1] != 'X' && dp1[j - 1][i])
can1 = true;
}
bool can2 = false;
if(right == n){
if(i == k - 1)
can2 = true;
}
else{
if(s[right] != 'X' && dp2[right + 1][k - (i + 1)])
can2 = true;
}
if(can1 && can2 && can3){
pp = max(pp, j);
while(pp < right){
canx[pp] = true;
pp++;
}
}
}
}
REP(i, n){
REP(j, k + 1){
if(dp1[i][j] && dp2[i + 1][(k - j)])
cany[i] = true;
}
}
REP(i, n){
if(s[i] == 'X')
ans[i] = 'X';
else if(s[i] == '_')
ans[i] = '_';
else{
if(canx[i] && !cany[i])
ans[i] = 'X';
else if(cany[i] && !canx[i])
ans[i] = '_';
}
}
return ans;
}
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