제출 #211608

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
211608		ryansee	Building 4 (JOI20_building4)	C++14	100 / 100	361 ms	18984 KiB

building4

#include "bits/stdc++.h"
using namespace std;

#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define eb emplace_back
#define ins insert
#define ph push
#define f first
#define s second
#define cbr cerr << "hi\n"
#define mmst(x, v) memset((x), v, sizeof ((x)))
#define siz(x) ll(x.size())
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
inline long long rand(long long x, long long y) { return (rng() % (y+1-x)) + x; } //inclusivesss
string inline to_string(char c) {string s(1,c);return s;} template<typename T> inline T gcd(T a,T b){ return a==0?llabs(b):gcd(b%a,a); }

typedef long long ll; 
typedef long double ld;
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
typedef pair<ll,ll>pi; typedef pair<ll,pi>spi; typedef pair<pi,pi>dpi;

#define LLINF ((long long)1e18)
#define INF int(1e9+1e6)
#define MAXN (1000006)
ll n,A[2][MAXN];
bool ans[MAXN],block[MAXN];
int main(){
	FAST
	cin>>n;
	n*=2;
	FOR(j,0,1)FOR(i,1,n)cin>>A[j][i];
	ans[1]=(A[0][1]<A[1][1]?0:1);
	FOR(i,2,n){
		if((A[0][i]<A[1][i]&&A[0][i]>=A[ans[i-1]][i-1])||(A[1][i]<A[ans[i-1]][i-1])){
			ans[i]=0;
		}else ans[i]=1;
		if(A[ans[i]][i]<A[ans[i-1]][i-1]){
			cout<<"-1\n";
			return 0;
		}
	}
	vector<ll> cnt(2,0);
	FOR(i,1,n)++cnt[ans[i]];
	FOR(i,1,n){
		if(cnt[0]==cnt[1])break;
		if(cnt[ans[i]]<cnt[!ans[i]])continue;
		if(A[!ans[i]][i]<A[ans[i]][i])continue;
		if(block[i])continue;
		ll cur=1, p=A[!ans[i]][i], j=i+1;
		bool gg=0;
		while(j<=n){
			if(A[ans[j]][j]>=p) break;
			if(A[!ans[j]][j]<p) { FOR(k,i,j-1) block[k]=1; gg=1; break; }
			if(cnt[ans[j]]>cnt[!ans[j]]) ++ cur;
			else -- cur;
			if(cur<0){FOR(k,i,j)block[k]=1;gg=1;break;}
			// if(cnt[!ans[i]]+cur>cnt[ans[i]]-cur){
				// gg=1;break;
			// }
			p=A[!ans[j]][j];
			++ j;
		} // j is where we cannot reach
		if(gg){
			continue;
		}
		if(cnt[!ans[i]]+cur>cnt[ans[i]]-cur){
			ll start=-1;
			FOR(k,i,j-1) {
				if(cnt[ans[k]]>cnt[!ans[k]]) -- cur; else ++ cur;
				if(cnt[!ans[i]]+cur==cnt[ans[i]]-cur){
					start=k+1;
				}
			}
			assert(~start);
			FOR(k,start,j-1)--cnt[ans[k]],ans[k]^=1,++cnt[ans[k]];
			i=j-1;
			continue;
		}
		// flip
		FOR(k,i,j-1)--cnt[ans[k]],ans[k]^=1,++cnt[ans[k]];
		i=j-1;
	}
	if(cnt[0]==cnt[1]){
		FOR(i,1,n)cout<<(ans[i]?'B':'A');
		cout<<'\n';
	}else cout<<"-1\n";
}

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