結果

問題 No.386 貪欲な領主
ユーザー torus711torus711
提出日時 2016-07-01 23:03:51
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 353 ms / 2,000 ms
コード長 5,483 bytes
コンパイル時間 1,149 ms
コンパイル使用メモリ 114,540 KB
実行使用メモリ 38,460 KB
最終ジャッジ日時 2023-08-02 08:55:35
合計ジャッジ時間 3,614 ms
ジャッジサーバーID
(参考情報)
judge12 / judge15
このコードへのチャレンジ(β)

テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 2 ms
4,380 KB
testcase_01 AC 2 ms
4,376 KB
testcase_02 AC 1 ms
4,380 KB
testcase_03 AC 2 ms
4,376 KB
testcase_04 AC 353 ms
38,460 KB
testcase_05 AC 264 ms
32,820 KB
testcase_06 AC 267 ms
32,808 KB
testcase_07 AC 3 ms
4,376 KB
testcase_08 AC 27 ms
5,992 KB
testcase_09 AC 4 ms
4,380 KB
testcase_10 AC 2 ms
4,376 KB
testcase_11 AC 2 ms
4,380 KB
testcase_12 AC 2 ms
4,376 KB
testcase_13 AC 7 ms
4,412 KB
testcase_14 AC 266 ms
32,980 KB
testcase_15 AC 282 ms
38,108 KB
権限があれば一括ダウンロードができます

ソースコード

diff #

#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 ) begin( c ), end( c )
#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 EXIST( c, e ) ( ( c ).find( e ) != ( c ).end() )

template < typename T > inline bool chmin( T &a, const T &b ){ if ( b < a ) { a = b; return true; } return false; }
template < typename T > inline bool chmax( T &a, const T &b ){ if ( a < b ) { a = b; return true; } return false; }

#define PB push_back
#define EM emplace
#define EB emplace_back
#define BI back_inserter

#define MP make_pair
#define fst first
#define snd second

#define DUMP( x ) cerr << #x << " = " << ( x ) << endl

// BEGIN SNIPLATE  lca
// {{ abbr: calculate Lowest Common Ancestor of a tree }}
// {{ class: Graph, Tree }}
// {{ pattern: class\sLowestCommonAncestor }}
// 木の根を適当に固定して、最近共通祖先を求める
// 二点間距離もいける
// 前処理 O( N log N )
// クエリ O( log N )
class LowestCommonAncestor
{
private:
	const int N;
	std::vector< VPII > G;

	bool constructed = false;
	std::vector<int> depth_, distances_;
	std::vector< std::vector<int> > parent_;

public:
	LowestCommonAncestor( const int n ) : N( n ), G( N ), depth_( N ), distances_( N ), parent_( N, std::vector<int>( 31, -1 ) )
	{
		return;
	}

	void connect( const int u, const int v, const int d = 1 )
	{
		G[u].EB( v, d );
		G[v].EB( u, d );
		constructed = false;

		return;
	}

	int depth( const int u ) const
	{
		return depth_[u];
	}

	int solve( int u, int v )
	{
		if ( !constructed )
		{
			dfs( 0, -1, 0, 0 );
			doubling();
			constructed = true;
		}

		if ( depth_[u] < depth_[v] )
		{
			swap( u, v );
		}

		{ // align depths
			int diff = depth_[u] - depth_[v];
			for ( int i = 30; 0 <= i; --i )
			{
				if ( diff & 1 << i )
				{
					u = parent_[u][i];
				}
			}
		}

		if ( u == v )
		{
			return u;
		}

		for ( int i = 30; 0 <= i; --i )
		{
			if ( parent_[u][i] != parent_[v][i] )
			{
				u = parent_[u][i];
				v = parent_[v][i];
			}
		}

		return parent_[u][0];
	}

	int distance( const int u, const int v )
	{
		const int w = solve( u, v );
		return distances_[u] + distances_[v] - 2 * distances_[w];
	}

private:
	void dfs( const int u, const int prev, const int d, const int dist )
	{
		depth_[u] = d;
		parent_[u][0] = prev;
		distances_[u] = dist;

		for ( const PII &e : G[u] )
		{
			const int v = e.fst;
			if ( v == prev )
			{
				continue;
			}

			dfs( v, u, d + 1, dist + e.snd );
		}

		return;
	}

	void doubling()
	{
		for ( int i = 0; i < 30; ++i )
		{
			for ( int u = 0; u < N; ++u )
			{
				if ( parent_[u][i] != -1 )
				{
					parent_[u][ i + 1 ] = parent_[ parent_[u][i] ][i];
				}
			}
		}
		return;
	}
};
// LowestCommonAncestor( N )
// connect( u, v [, dist = 1 ] )
// solve( u, v )
// depth( u )
// distance( u, v )

int N;
VVI G;

VI costs, tcosts;

void dfs( const int u, const int prev = -1, const int cost = 0 )
{
	tcosts[u] = cost + costs[u];

	FOR( v, G[u] )
	{
		if ( v == prev )
		{
			continue;
		}
		dfs( v, u, cost + costs[u] );
	}

	return;
}

int main()
{
	cin.tie( 0 );
	ios::sync_with_stdio( false );
	cout << setprecision( 12 ) << fixed;

	scanf( "%d", &N );
	G.resize( N );
	costs.resize( N );
	tcosts.resize( N );

	LowestCommonAncestor lca( N );

	REP( _, N - 1 )
	{
		int u, v;
		scanf( "%d%d", &u, &v );

		G[u].PB( v );
		G[v].PB( u );

		lca.connect( u, v );
	}

	for_each( ALL( costs ), []( int &c ){ scanf( "%d", &c ); } );

	dfs( 0 );

	int M;
	scanf( "%d", &M );

	LL res = 0;
	REP( _, M )
	{
		int a, b, c;
		scanf( "%d%d%d", &a, &b, &c );

		res -= tcosts[ lca.solve( a, b ) ] * c * 2;
		res += costs[ lca.solve( a, b ) ] * c;
		res += tcosts[a] * c;
		res += tcosts[b] * c;
	}

	printf( "%lld\n", res );

	return 0;
}
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