이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <math.h>
#include <cstdio>
#include <stdio.h>
#include <queue>
#include <cstdlib>
#include <deque>
#include <cassert>
#include "paint.h"
using namespace std;
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
#define mp make_pair
#define f first
#define se second
#define pb push_back
#define ppb pop_back
#define ll long long
#define ull unsigned long long
#define cntbit(x) __builtin_popcount(x)
#define endl '\n'
#define uset unordered_set
#define umap unordered_map
#define pii pair<int, int>
#define ld long double
#define pll pair<long long, long long>
const int N = 5e5 + 15;
const int K = 1e2 + 15;
int n, k, p[2][N], a[K], pref[2][N];
bool dp[2][K][N];
inline void calc(string &s) {
p[0][0] = p[1][0];
for(int i = 1; i <= n; ++i)
p[0][i] = p[0][i-1] + (s[i] == '_'),
p[1][i] = p[1][i-1] + (s[i] == 'X');
}
inline void calc_dp(string &s, bool dp[K][N]) {
dp[0][0] = true;
for(int i = 0; i <= k; ++i) {
for(int j = 1; j <= n; ++j) {
if(s[j] != '_') {
if(i > 0 && j >= a[i] && p[0][j] == p[0][j - a[i]] && s[j - a[i]] != 'X')
dp[i][j] |= dp[i-1][max(0, j - a[i] - 1)];
}
if(s[j] != 'X')
dp[i][j] |= dp[i][j-1];
}
}
}
string solve_puzzle(string s, vector<int> c) {
string ans = "";
n = s.size(); k = c.size();
s = '#' + s + '#';
for(int i = 0; i < k; ++i)
a[i+1] = c[i];
calc(s);
calc_dp(s, dp[0]);
reverse(a + 1, a + 1 + k);
reverse(s.begin(), s.end());
calc(s);
calc_dp(s, dp[1]);
reverse(dp[1], dp[1] + k + 2);
reverse(a + 1, a + 1 + k);
reverse(s.begin(), s.end());
calc(s);
for(int i = 0; i <= k + 1; ++i)
reverse(dp[1][i], dp[1][i] + n + 2);
for(int i = 0; i <= k; ++i)
for(int j = 1; j <= n; ++j)
if(dp[0][i][j-1] && s[j] != 'X' && dp[1][i+1][j+1])
++pref[0][j];
for(int i = 1; i <= k; ++i) {
for(int j = 1; j + a[i] - 1 <= n; ++j) {
if(p[0][j + a[i] - 1] == p[0][j-1] && s[j-1] != 'X' && s[j + a[i]] != 'X' && dp[1][i+1][min(n + 1, j + a[i] + 1)] && dp[0][i-1][max(j-2, 0)]) {
++pref[1][j];
--pref[1][j + a[i]];
}
}
}
for(int i = 1; i <= n; ++i) {
pref[1][i] += pref[1][i-1];
if(pref[1][i] && pref[0][i])
ans += '?';
else if(pref[1][i])
ans += 'X';
else
ans += '_';
}
return ans;
}
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