이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
ID: varunra2
LANG: C++
TASK: building4
*/
#include <bits/stdc++.h>
using namespace std;
#ifdef DEBUG
#include "lib/debug.h"
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define debug_arr(...) \
cerr << "[" << #__VA_ARGS__ << "]:", debug_arr(__VA_ARGS__)
#pragma GCC diagnostic ignored "-Wsign-compare"
//#pragma GCC diagnostic ignored "-Wunused-parameter"
//#pragma GCC diagnostic ignored "-Wunused-variable"
#else
#define debug(...) 42
#endif
#define EPS 1e-9
#define IN(A, B, C) assert(B <= A && A <= C)
#define INF (int)1e9
#define MEM(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define MP make_pair
#define PB push_back
#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.end(), cont.begin()
#define x first
#define y second
const double PI = acos(-1.0);
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef map<int, int> MPII;
typedef multiset<int> MSETI;
typedef set<int> SETI;
typedef set<string> SETS;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef vector<string> VS;
#define rep(i, a, b) for (int i = a; i < (b); ++i)
#define trav(a, x) for (auto& a : x)
#define sz(x) (int)(x).size()
typedef pair<int, int> pii;
typedef vector<int> vi;
#pragma GCC diagnostic ignored "-Wsign-compare"
// util functions
int n;
void solve(VVI& dpl, VVI& dpr, int i, int j, VI& ret, int cnt, VVI& vals) {
ret[i] = j;
if (j == 0) cnt++;
if (i == 0) return;
if (dpl[i - 1][0] <= (n - cnt) and dpr[i - 1][0] >= (n - cnt) and
vals[0][i - 1] <= vals[j][i]) {
solve(dpl, dpr, i - 1, 0, ret, cnt, vals);
} else {
solve(dpl, dpr, i - 1, 1, ret, cnt, vals);
}
}
int main() {
cin.sync_with_stdio(0);
cin.tie(0);
cin >> n;
VVI vals(2, VI(2 * n));
trav(x, vals[0]) cin >> x;
trav(x, vals[1]) cin >> x;
VVI dpl(2 * n, VI(2, INF));
VVI dpr(2 * n, VI(2, -1));
dpl[0][0] = dpr[0][0] = 1;
dpl[0][1] = dpr[0][1] = 0;
for (int i = 1; i < 2 * n; i++) {
for (int j = 0; j < 2; j++) {
for (int k = 0; k < 2; k++) {
if (dpl[i - 1][k] == INF) continue;
if (vals[j][i] < vals[k][i - 1]) continue;
// you can go from (i - 1, k) -> (i, j)
dpl[i][j] = min(dpl[i][j], dpl[i - 1][k] + (j == 0 ? 1 : 0));
dpr[i][j] = max(dpr[i][j], dpr[i - 1][k] + (j == 0 ? 1 : 0));
}
}
}
VI ret(2 * n, 0);
if (dpl.back()[0] > n or dpr.back()[0] < n) {
if (dpl.back()[1] > n or dpr.back()[1] < n) {
cout << "-1\n";
return 0;
} else {
solve(dpl, dpr, 2 * n - 1, 1, ret, 0, vals);
}
} else {
solve(dpl, dpr, 2 * n - 1, 0, ret, 0, vals);
}
for (int i = 0; i < 2 * n; i++) {
if (ret[i] == 0)
cout << 'A';
else
cout << 'B';
}
cout << '\n';
return 0;
}
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