結果
問題 | No.258 回転寿司(2) |
ユーザー |
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提出日時 | 2015-07-31 22:37:08 |
言語 | C++11 (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 2,984 bytes |
コンパイル時間 | 961 ms |
コンパイル使用メモリ | 96,612 KB |
実行使用メモリ | 5,248 KB |
最終ジャッジ日時 | 2024-11-06 18:43:34 |
合計ジャッジ時間 | 4,266 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 4 |
other | AC * 67 |
ソースコード
#include <iostream>#include <iomanip>#include <sstream>#include <vector>#include <string>#include <set>#include <unordered_set>#include <map>#include <unordered_map>#include <stack>#include <queue>#include <deque>#include <algorithm>#include <functional>#include <iterator>#include <limits>#include <numeric>#include <utility>#include <cmath>#include <cassert>#include <cstdio>using namespace std; using namespace placeholders;using LL = long long;using ULL = unsigned long long;using VI = vector< int >;using VVI = vector< vector< int > >;using VS = vector< string >;using SS = stringstream;using PII = pair< int, int >;using VPII = vector< pair< int, int > >;template < typename T = int > using VT = vector< T >;template < typename T = int > using VVT = vector< vector< T > >;template < typename T = int > using LIM = numeric_limits< T >;template < typename T > inline istream& operator>>( istream &s, vector< T > &v ){ for ( T &t : v ) { s >> t; } return s; }template < typename T > inline ostream& operator<<( ostream &s, const vector< T > &v ){ for ( int i = 0; i < int( v.size() ); ++i ){ s << ( " " + !i) << v[i]; } return s; }template < typename T > inline T fromString( const string &s ) { T res; istringstream iss( s ); iss >> res; return res; };template < typename T > inline string toString( const T &a ) { ostringstream oss; oss << a; return oss.str(); };#define REP2( i, n ) REP3( i, 0, n )#define REP3( i, m, n ) for ( int i = ( int )( m ); i < ( int )( n ); ++i )#define GET_REP( a, b, c, F, ... ) F#define REP( ... ) GET_REP( __VA_ARGS__, REP3, REP2 )( __VA_ARGS__ )#define FOR( e, c ) for ( auto &e : c )#define ALL( c ) ( c ).begin(), ( c ).end()#define AALL( a, t ) ( t* )a, ( t* )a + sizeof( a ) / sizeof( t )#define DRANGE( c, p ) ( c ).begin(), ( c ).begin() + ( p ), ( c ).end()#define SZ( v ) ( (int)( v ).size() )#define PB push_back#define EM emplace#define EB emplace_back#define BI back_inserter#define EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )#define MP make_pair#define fst first#define snd second#define DUMP( x ) cerr << #x << " = " << ( x ) << endlconstexpr int INF = LIM<>::max() / 2;int dp[ 1024 ][2];int prev_i[ 1024 ][2];int prev_j[ 1024 ][2];int main(){cin.tie( 0 );ios::sync_with_stdio( false );int N;cin >> N;VI V( N );cin >> V;fill( AALL( dp, int ), -INF );dp[0][0] = 0;REP( i, N ){if ( dp[ i + 1 ][1] < dp[i][0] + V[i] ){dp[ i + 1 ][1] = dp[i][0] + V[i];prev_i[ i + 1 ][1] = i;prev_j[ i + 1 ][1] = 0;}REP( j, 2 ){if ( dp[ i + 1 ][0] < dp[i][j] ){dp[ i + 1 ][0] = dp[i][j];prev_i[ i + 1 ][0] = i;prev_j[ i + 1 ][0] = j;}}}VI res;for ( int i = N, j = dp[N][0] < dp[N][1]; 0 < i; tie( i, j ) = tie( prev_i[i][j], prev_j[i][j] ) ){if ( j ){res.PB( i );}}reverse( ALL( res ) );cout << max( dp[N][0], dp[N][1] ) << endl;cout << res << endl;return 0;}