Submission #66874

#TimeUsernameProblemLanguageResultExecution timeMemory
66874BenqSailing Race (CEOI12_race)C++14
75 / 100
3060 ms4832 KiB
#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 N,k; int nor(int x) { x = (x%N+N)%N; if (x == 0) x += N; return x; } template<class T> void MX(T& a, T b) { a = max(a,b); } int alt[2][501][501]; pi solve0() { FOR(len,2,N) FOR(i,1,N+1) { int en = nor(i+len); FOR(j,1,len) { int m = nor(i+j); int ex = max(alt[0][m][en],alt[1][i][m])+1; if (ok[i][m]) MX(alt[0][i][en],ex); if (ok[en][m]) MX(alt[1][i][en],ex); } } pi ret = {0,0}; FOR(i,1,N+1) FOR(j,1,N+1) if (ok[i][j]) MX(ret,{max(alt[1][i][j],alt[0][j][i])+1,i}); return ret; } int no[2][501][501]; pi ret = {0,0}; void test1(int len, int i) { pi yes = {-MOD,-MOD}; int en = nor(i+len); int existsEdge = 0; for (int j = len-1; j >= 0; --j) { int m = nor(i+j); if (ok[m][en]) { MX(no[0][i][en],no[0][i][m]+1); if (existsEdge) MX(yes,{no[0][i][en],existsEdge}); } if (ok[m][i]) existsEdge = m; } MX(ret,{yes.f+alt[0][en][i]+1,yes.s}); } void test2(int len, int i) { int en = nor(i+len); no[0][i][en] = -MOD; for (int j = 0; j < len; ++j) { int m = nor(i+j); if (j && ok[m][i]) MX(ret,{no[0][i][en]+alt[1][m][en]+1,m}); if (ok[m][en]) MX(no[0][i][en],no[0][i][m]+1); } } void test3(int len, int i) { pi yes = {-MOD,-MOD}; int st = nor(i-len); int existsEdge = 0; for (int j = len-1; j >= 0; --j) { int m = nor(i-j); if (ok[m][st]) { MX(no[1][st][i],no[1][m][i]+1); if (existsEdge) MX(yes,{no[1][st][i],existsEdge}); } if (ok[m][i]) existsEdge = m; } // cout << "OH " << yes.f+alt[1][i][st]+1 << " " << yes.f << " " << yes.s << " " << i << " " << st << "\n"; MX(ret,{yes.f+alt[1][i][st]+1,yes.s}); } void test4(int len, int i) { int st = nor(i-len); no[1][st][i] = -MOD; for (int j = 0; j < len; ++j) { int m = nor(i-j); if (j && ok[m][i]) MX(ret,{no[1][st][i]+alt[0][st][m]+1,m}); if (ok[m][st]) MX(no[1][st][i],no[1][m][i]+1); } } pi solve1() { F0R(i,2) FOR(j,1,N+1) FOR(k,1,N+1) { if (j != k) no[i][j][k] = -MOD; else no[i][j][k] = 0; } FOR(len,1,N) FOR(i,1,N+1) { test1(len,i); test2(len,i); test3(len,i); test4(len,i); } 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) MX(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 */
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