제출 #295875

#제출 시각아이디문제언어결과실행 시간메모리
295875eggag32건물 4 (JOI20_building4)C++17
11 / 100
27 ms5748 KiB
#pragma GCC optimize ("O3") #pragma GCC target ("sse4") #include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef vector<int> vi; typedef pair<int, int> pi; #define debug(x) cerr << #x << ": " << x << endl #define debug2(x, y) debug(x), debug(y) #define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++) #define rep(i, a) for(int i = 0; i < (int)(a); i++) #define all(v) v.begin(), v.end() #define mp make_pair #define pb push_back #define lb lower_bound #define ub upper_bound #define fi first #define se second #define sq(x) ((x) * (x)) const int mxN = 2e5 + 5; template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); } int a[mxN][2]; int mx[mxN][2], mn[mxN][2]; int main(){ ios_base::sync_with_stdio(false); cin.tie(0); //freopen("input.in", "r", stdin); //freopen("output.out", "w", stdout); int n; cin >> n; n *= 2; rep(j, 2) repn(i, 1, n + 1) cin >> a[i][j]; repn(i, 1, n + 1){ rep(j, 2){ //what we take mn[i][j] = 1e9, mx[i][j] = -1e9; rep(k, 2){ //what we update from if(mn[i - 1][k] < 0 || a[i][j] < a[i - 1][k]) continue; mn[i][j] = min(mn[i][j], mn[i - 1][k] + (j == 0)); mx[i][j] = max(mx[i][j], mx[i - 1][k] + (j == 0)); } } } vi ans; if(mn[n][0] <= n / 2 && mx[n][0] >= n / 2) ans.pb(0); else if(mn[n][1] <= n / 2 && mx[n][1] >= n / 2) ans.pb(1); else{ cout << -1 << endl; return 0; } int len = n / 2; for(int i = n - 1; i >= 1; i--){ int lst = ans.back(); if(lst == 0) len--; rep(j, 2){ if(mx[i][j] < 0 || a[i + 1][lst] < a[i][j]) continue; if(mn[i][j] <= len && mx[i][j] >= len){ ans.pb(j); break; } } } for(int i = n - 1; i >= 0; i--) cout << (char)('A' + ans[i]); cout << endl; return 0; } /* Things to look out for: - Integer overflows - Array bounds - Special cases Be careful! */
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