Submission #62908

#	Time	Username	Problem	Language	Result	Execution time	Memory
62908		Benq	Sailing Race (CEOI12_race)	C++11	75 / 100	3061 ms	5752 KiB

race

#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;

bool ok[501][501];
int dp[2][2][501][501];
int N,k;

int nor(int x) { x = (x%N+N)%N; if (x == 0) x += N; return x; }

void MX(int& a, int b) { a = max(a,b); }

pi solve0() {
    FOR(len,2,N) FOR(i,1,N+1) {
        int en = nor(i+len);
        FOR(j,1,len) {
            int ex = max(dp[0][0][nor(i+j)][en],dp[0][1][i][nor(i+j)])+1;
            if (ok[i][nor(i+j)]) MX(dp[0][0][i][en],ex);
            if (ok[en][nor(i+j)]) MX(dp[0][1][i][en],ex);
        }
    }
    pi ret = {0,0};
    FOR(i,1,N+1) FOR(j,1,N+1) if (ok[i][j]) 
        ret = max(ret,{max(dp[0][1][i][j],dp[0][0][j][i])+1,i});
    return ret;
}

pi solve1() {
    FOR(len,1,N) FOR(i,1,N+1) {
        int en = nor(i+len);
        dp[1][1][i][en] = dp[1][0][i][en] = -MOD;
        F0R(j,len+1) {
            if (ok[nor(i+j)][en]) MX(dp[1][1][i][en],dp[1][1][i][nor(i+j)]+1);
            if (ok[nor(i+j)][i]) MX(dp[1][0][i][en],dp[1][0][nor(i+j)][en]+1);
        }
    }
    pi ret = {0,0};
    FOR(a,1,N+1) FOR(len,1,N+1) if (ok[a][nor(a+len)]) {
        vi l, r;
        FOR(b,1,N+1) if (b != len) {
            if (b < len) l.pb(nor(a+b));
            else r.pb(nor(a+b));
        } 
        for (int i: l) for (int j: r) if (ok[i][j]) 
            ret = max(ret,{1+dp[1][0][i][nor(a+len)]+1+max(dp[0][0][j][a],dp[0][1][nor(a+len)][j]),a});
        for (int i: r) for (int j: l) if (ok[i][j])
            ret = max(ret,{1+dp[1][1][nor(a+len)][i]+1+max(dp[0][0][j][nor(a+len)],dp[0][1][a][j]),a});
    }
    return ret;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> k;
    FOR(i,1,N+1) {
        int x; 
        while (cin >> x) {
            if (x == 0) break;
            ok[i][x] = 1;
        }
    }
    pi t = solve0();
    if (k == 1) t = max(t,solve1());
    cout << t.f << "\n" << t.s;
}

/* 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: 75 / 100