Submission #574074

#	Time	Username	Problem	Language	Result	Execution time	Memory
574074		MohammadAghil	Building 4 (JOI20_building4)	C++14	100 / 100	279 ms	45144 KiB

building4

      #include <bits/stdc++.h>
//   #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
    using namespace std;
  typedef long long ll;
   typedef pair<int, int> pp;
    #define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
      #define per(i,r,l) for(int i = (r); i >= (l); i--)
        #define rep(i,l,r) for(int i = (l); i < (r); i++)
           #define all(x) x.begin(), x.end()
              #define sz(x) (int)(x).size()
                  #define pb push_back
                      #define ss second
                           #define ff first
                                   void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 1e6 + 5, lg = 21, inf = ll(1e9) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;}

int a[maxn], b[maxn], dpmx[maxn][2], dpmn[maxn][2];

int main(){
     cin.tie(0) -> sync_with_stdio(0);
     int n; cin >> n;
     rep(i,0,n<<1) cin >> a[i];
     rep(i,0,n<<1) cin >> b[i];
     rep(i,0,2) dpmx[0][i] = dpmn[0][i] = i;
     rep(i,1,n<<1){
          rep(j,0,2) dpmn[i][j] = (n<<1)+1;
          if(a[i] >= a[i-1]) dpmx[i][1] = max(dpmx[i][1], dpmx[i-1][1]+1), dpmn[i][1] = min(dpmn[i][1], dpmn[i-1][1]+1);
          if(a[i] >= b[i-1]) dpmx[i][1] = max(dpmx[i][1], dpmx[i-1][0]+1), dpmn[i][1] = min(dpmn[i][1], dpmn[i-1][0]+1);
          if(b[i] >= a[i-1]) dpmx[i][0] = max(dpmx[i][0], dpmx[i-1][1]), dpmn[i][0] = min(dpmn[i][0], dpmn[i-1][1]);
          if(b[i] >= b[i-1]) dpmx[i][0] = max(dpmx[i][0], dpmx[i-1][0]), dpmn[i][0] = min(dpmn[i][0], dpmn[i-1][0]);
          // er(i, dpmx[i][0], dpmx[i][1], dpmn[i][0], dpmn[i][1]);
     }
     int en = (n<<1)-1;
     if((dpmx[en][0] < n || dpmn[en][0] > n) && (dpmx[en][1] < n || dpmn[en][1] > n)) return cout << "-1\n", 0;
     bool cr = false;
     if(dpmx[en][0] < n || dpmn[en][0] > n) cr = true;
     string ans(n<<1,'A');
     int cnt = 0;
     per(i,en,0){
          ans[i] = cr?'A':'B', cnt += cr;
          if(i){
               if(cnt + dpmx[i-1][cr] < n || cnt + dpmn[i-1][cr] > n || (cr?a[i]<a[i-1]:b[i]<b[i-1])) cr ^= 1;
          }
     }
     cout << ans << '\n';
     return 0; 
}

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