Submission #320782

#	Time	Username	Problem	Language	Result	Execution time	Memory
320782		aZvezda	Building 4 (JOI20_building4)	C++14	11 / 100	120 ms	105964 KiB

building4

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#define endl "\n"
typedef long long ll;
template<class T, class T2> inline ostream &operator <<(ostream &out, const pair<T, T2> &x) { out << x.first << " " << x.second; return out;}
template<class T, class T2> inline istream &operator >>(istream &in, pair<T, T2> &x) { in >> x.first >> x.second; return in;}
template<class T, class T2> inline bool chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline bool chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const ll mod = 1e9 + 7;
#define out(x) "{" << (#x) << ": " << x << "} "

const int MAX_N = 4e3 + 10;
int n, a[MAX_N], b[MAX_N];
int dp[MAX_N][MAX_N][2];
bool hist[MAX_N][MAX_N][2];

void output(int x, int y, int last) {
	if(x == -1) {return;}
	output(x - 1, y - last, hist[x][y][last]);
	if(last) {
		cout << "B";
	} else {
		cout << "A";
	}
}

signed main() {
	ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	cin >> n;
	n *= 2;
	for(int i = 0; i < n; i ++) {
		cin >> a[i];
	}
	for(int i = 0; i < n; i ++) {
		cin >> b[i];
	}
	dp[0][0][0] = 1;
	dp[0][1][1] = 1;
	for(int i = 0; i < n; i ++) {
		for(int j = 0; j < n; j ++) {
			if(dp[i][j][0]) {
				int now = a[i];
				if(now <= a[i + 1]) {
					dp[i + 1][j][0] = 1;
					hist[i + 1][j][0] = 0;
				} 
				if(now <= b[i + 1]) {
					dp[i + 1][j + 1][1] = 1;
					hist[i + 1][j + 1][1] = 0;
				}
			}
			if(dp[i][j][1]) {
				int now = b[i];
				if(now <= a[i + 1]) {
					dp[i + 1][j][0] = 1;
					hist[i + 1][j][0] = 1;
				} 
				if(now <= b[i + 1]) {
					dp[i + 1][j + 1][1] = 1;
					hist[i + 1][j + 1][1] = 1;
				}
			}
		}
	}
	
	if(!dp[n - 1][n / 2][0] && !dp[n - 1][n / 2][1]) {
		cout << -1 << endl;
	} else if(dp[n - 1][n / 2][1]) {
		output(n - 1, n / 2, 1);
	} else {
		output(n - 1, n / 2, 0);
	}
	return 0;
	
	for(int i = 0; i < n; i ++) {
		for(int j = 0; j < n; j ++) {
			cout << dp[i][j][0] << " ";
		}
		cout << endl;
	}
	cout << endl;
	for(int i = 0; i < n; i ++) {
		for(int j = 0; j < n; j ++) {
			cout << dp[i][j][1] << " ";
		}
		cout << endl;
	}
	
	return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 0 / 89