This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#include "paint.h"
#ifdef MINA
#include "grader.cpp"
#endif
using namespace std;
#define ll long long
const int N = 200'005, K = 105;
bool prfdp[N][K], sufdp[N][K];
string solve_puzzle(string s, vector<int> c) {
int n = s.size(), k = c.size();
int prf[n + 1]{};
for (int i = 0; i < n; i++) {
prf[i + 1] = prf[i] + (s[i] == '_');
}
prfdp[0][0] = 1;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= k; j++) {
if (j - 1 >= 0) {
if (i - c[j - 1] >= 0 && prf[i] - prf[i - c[j - 1]] == 0) {
if (j - 1 == 0) {
prfdp[i][j] = prfdp[i - c[j - 1]][j - 1];
} else if (s[i - c[j - 1] - 1] != 'X' && i - c[j - 1] - 1 >= 0) {
prfdp[i][j] = prfdp[i - c[j - 1] - 1][j - 1];
}
}
}
if (s[i - 1] != 'X') {
prfdp[i][j] |= prfdp[i - 1][j];
}
}
}
sufdp[n][k] = 1;
for (int i = n - 1; i >= 0; i--) {
for (int j = 0; j <= k; j++) {
if (j + 1 <= k) {
if (i + c[j] <= n && prf[i + c[j]] - prf[i] == 0) {
if (j + 1 == k) {
sufdp[i][j] = sufdp[i + c[j]][j + 1];
} else if (i + c[j] + 1 <= n && s[i + c[j]] != 'X') {
sufdp[i][j] = sufdp[i + c[j] + 1][j + 1];
}
}
}
if (s[i] != 'X') {
sufdp[i][j] |= sufdp[i + 1][j];
}
}
}
for (int j = 0; j <= k; j++) {
int cur = 0, cnt = 0;
for (int i = 1; i <= n; i++) {
if (prfdp[i][j]) {
cur++;
} else {
if (cur) {
cnt++;
}
cur = 0;
}
}
assert(cnt <= 1);
}
for (int j = 0; j <= k; j++) {
int cur = 0, cnt = 0;
for (int i = 0; i < n; i++) {
if (sufdp[i][j]) {
cur++;
} else {
if (cur) {
cnt++;
}
cur = 0;
}
}
assert(cnt <= 1);
}
string ans;
ans.resize(n);
for (int i = 0; i < n; i++) {
if (s[i] != '.') {
ans[i] = s[i];
continue;
}
bool ok[2]{};
for (int j = 0; j < k; j++) {
for (int l = i; l >= max(0, i - c[j] + 1); l--) {
if (l + c[j] <= n && prf[l + c[j]] - prf[l] == 0) {
bool vl = 0, vr = 0;
if (j == 0) {
vl = prfdp[l][j];
} else if (l - 1 >= 0 && s[l - 1] != 'X') {
vl = prfdp[l - 1][j];
}
if (j + 1 == k) {
vr = sufdp[l + c[j]][j + 1];
} else if (l + c[j] + 1 <= n && s[l + c[j]] != 'X') {
vr = sufdp[l + c[j] + 1][j + 1];
}
ok[1] |= vl && vr;
}
}
}
for (int j = 0; j <= k; j++) {
ok[0] |= prfdp[i][j] && sufdp[i + 1][j];
}
if (ok[0] && ok[1]) {
ans[i] = '?';
} else if (ok[0]) {
ans[i] = '_';
} else if (ok[1]) {
ans[i] = 'X';
} else {
assert(0);
}
}
return ans;
}
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