Submission #66483

#	Time	Username	Problem	Language	Result	Execution time	Memory
66483		Benq	Round words (IZhO13_rowords)	C++11	100 / 100	129 ms	26876 KiB

rowords

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

pi dp[2001][4001];
string A,B;
 
void init() {
    FOR(i,1,sz(A)+1) FOR(j,1,sz(B)+1) {
        pi bes = {-1,-1};
        bes = max(bes,{dp[i-1][j].f,0});
        bes = max(bes,{dp[i-1][j-1].f+(A[i-1] == B[j-1]),-1});
        bes = max(bes,{dp[i][j-1].f,-2});
        dp[i][j] = {bes.f,-bes.s};
    }
}
 
void adjust(int col) {
    int x = 1;
    while (x <= sz(A) && dp[x][col].s == 0) x++;
    if (x > sz(A)) return;
    
    dp[x][col].s = 0;
    pi cur = {x,col};
    while (cur.f <= sz(A) && cur.s <= sz(B)) {
        if (cur.s < sz(B) && dp[cur.f][cur.s+1].s == 2) {
            cur.s ++;
            dp[cur.f][cur.s].s = 0;
        } else if (cur.f < sz(A) && cur.s < sz(B) && dp[cur.f+1][cur.s+1].s == 1) {
            cur.f ++, cur.s ++;
            dp[cur.f][cur.s].s = 0;
        } else cur.f ++;
    }
}

int get(pi x) {
    int lo = x.s-sz(B)/2;
    int ret = 0;
    while (x.f && x.s > lo) {
        if (dp[x.f][x.s].s == 0) x.f --;
        else if (dp[x.f][x.s].s == 1) ret ++, x.f --, x.s --;
        else x.s --;
    }
    return ret;
}

int solve(string a, string b) {
    A = a, B = b+b;
    init();
    int ans = 0;
    F0R(i,sz(b)) {
        ans = max(ans,get({sz(a),i+sz(b)}));
        adjust(i+1);
    }
    return ans;
}
 
int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    string a,b; cin >> a >> b;
    int ans = solve(a,b);
    reverse(all(a));
    ans = max(ans,solve(a,b));
    cout << ans;
}
 

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 100 / 100