Submission #722380

#	Time	Username	Problem	Language	Result	Execution time	Memory
722380		urosk	원형 문자열 (IZhO13_rowords)	C++14	100 / 100	1970 ms	53312 KiB

rowords

#define here cerr<<"===========================================\n"
#define dbg(x) cerr<<#x<<": "<<x<<endl;
#include "bits/stdc++.h"
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/pb_ds/assoc_container.hpp>
#define ld double
#define ll long long
#define llinf 100000000000000000LL // 10^17
#define pb push_back
#define popb pop_back
#define fi first
#define sc second
#define endl '\n'
#define pll pair<ll,ll>
#define pld pair<ld,ld>
#define all(a) a.begin(),a.end()
#define ceri(a,l,r) {cerr<<#a<<": ";for(ll i_ = l;i_<=r;i_++) cerr<<a[i_]<< " ";cerr<<endl;}
#define cer(a) {cerr<<#a<<": ";for(ll x_ : a) cerr<<x_<< " ";cerr<<endl;}

#define daj_mi_malo_vremena ios_base::sync_with_stdio(false);cerr.tie(0);cout.tie(0);cin.tie(0);

using namespace std;
//using namespace __gnu_pbds;
/*
ll add(ll x,ll y){
    x+=y;
    if(x<0){
        x%=mod;
        x+=mod;
    }else{
        if(x>=mod) x%=mod;
    }
    return x;
}
ll mul(ll a,ll b){
	ll ans = (a*b)%mod;
	if(ans<0) ans+=mod;
	return ans;
}
ll po(ll x,ll y){
    if(y==0) return 1LL;
    ll ans = po(x,y/2);
    ans = mul(ans,ans);
    if(y&1) ans = mul(ans,x);
    return ans;
}
ll inv(ll x){return po(x,mod-2);}
*/
/*
typedef tree<int,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
typedef tree<int,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_multiset;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
ll rnd(ll l,ll r){
    return uniform_int_distribution<ll>(l,r)(rng);
}
*/
#define maxn 4005
ll n,m;
ll dpl[maxn][maxn],dpu[maxn][maxn];
string a,b;
void upd(string a,string b,ll x,ll y){
    ll c = (a[x-1]==b[y-1])|dpu[x][y-1]|dpl[x-1][y];
    dpl[x][y] = c-dpu[x][y-1];
    dpu[x][y] = c-dpl[x-1][y];
}
ll reshi(string a,string b){
    b+=b;
    n = a.size();
    m = b.size();
    for(ll i = 0;i<=n;i++) for(ll j = 0;j<=m;j++) dpu[i][j] = dpl[i][j] = 0;
    for(ll i = 1;i<=n;i++) for(ll j = 1;j<=m;j++) upd(a,b,i,j);
    ll ans = 0;
    for(ll i = 1;i<=n;i++) ans+=dpu[i][m/2];
    for(ll j = 1;j<m/2;j++){
        for(ll x = 1,y = j;x<=n&&y<=m;){
            ll last = dpu[x][y];
            if(j==y) dpu[x][y] = 0;
            else upd(a,b,x,y);
            if(last==dpu[x][y]) x++;
            else y++;
        }
        ll cur = 0;
        for(ll i = 1;i<=n;i++) cur+=dpu[i][j+m/2];
        ans = max(ans,cur);
    }
    return ans;
}
void tc(){
    cin >> a >> b;
    ll ans = reshi(a,b);
    reverse(all(b));
    ans = max(ans,reshi(a,b));
    cout<<ans<<endl;
}
int main(){
	daj_mi_malo_vremena
    int t; t = 1;
    while(t--){
        tc();
    }
	return 0;
}
/**
algorithm
grammar
4
**/

• Subtask 1: 100 / 100