「TJOI2019」甲苯先生的滚榜

题目链接

问题分析

参照数据范围,我们需要一个能够在(O(nlog n))复杂度内维护有序数列的数据结构。那么平衡树是很好的选择。参考程序中使用带旋Treap。

参考程序

#pragma GCC optimize( 3 )
#include <cstdio>
#include <ctime>
#include <algorithm>

namespace Treap {
	struct member {
		int Number, Time;
		bool operator > ( const member Other ) const {
			return Number < Other.Number || Number == Other.Number && Time > Other.Time;
		};
		bool operator == ( const member Other ) const {
			return Number == Other.Number && Time == Other.Time;
		};
		bool operator < ( const member Other ) const {
			return Number > Other.Number || Number == Other.Number && Time < Other.Time;
		}
	};
	struct node {
		int Random, Size, Cnt;
		member Value;
		node *LeftChild, *RightChild;
	};
	void Collect( node *A ) {
		A->Size = A->Cnt + ( ( A->LeftChild != NULL ) ? A->LeftChild->Size : 0 ) + ( ( A->RightChild != NULL ) ? A->RightChild->Size : 0 );
		return;
	}
	node *LeftRotate( node *A ) {
		node *B = A->RightChild; A->RightChild = B->LeftChild; B->LeftChild = A; Collect( A ); Collect( B ); return B;
	}
	node *RightRotate( node *A ) {
		node *B = A->LeftChild; A->LeftChild = B->RightChild; B->RightChild = A; Collect( A ); Collect( B ); return B;
	}
	node *Insert( node *Rt, member x ) {
		if( Rt == NULL ) {
			Rt = new node; 
			Rt->Random = rand() % 1000000000; Rt->Value = x; Rt->Size = 1; Rt->Cnt = 1; Rt->LeftChild = Rt->RightChild = NULL;
			return Rt;
		}
		++( Rt->Size );
		if( Rt->Value == x ) { ++( Rt->Cnt ); return Rt; }
		if( Rt->Value < x ) { 
			Rt->RightChild = Insert( Rt->RightChild, x ); if( Rt->RightChild->Random < Rt->Random ) Rt = LeftRotate( Rt );
		} else {
			Rt->LeftChild = Insert( Rt->LeftChild, x ); if( Rt->LeftChild->Random < Rt->Random ) Rt = RightRotate( Rt );
		}
		return Rt;
	}
	node *Del( node *Rt, member x ) {
		if( Rt == NULL ) { printf( "No such number called %d
", x ); return Rt; }
		if( Rt->Value == x ) {
			if( Rt->Cnt > 1 ) { --( Rt->Cnt ); --( Rt->Size ); return Rt; }
			if( Rt->LeftChild == NULL ) { node *T = Rt->RightChild; delete Rt; return T; }
			if( Rt->RightChild == NULL ) { node *T = Rt->LeftChild; delete Rt; return T; }
			if( Rt->LeftChild->Random <= Rt->RightChild->Random ) {
				Rt = RightRotate( Rt ); --( Rt->Size ); Rt->RightChild = Del( Rt->RightChild, x ); return Rt;
			} else {
				Rt = LeftRotate( Rt ); --( Rt->Size ); Rt->LeftChild = Del( Rt->LeftChild, x ); return Rt;
			}
			return Rt;
		}
		--( Rt->Size );
		if( Rt->Value < x ) { Rt->RightChild = Del( Rt->RightChild, x ); return Rt; }
		else { Rt->LeftChild = Del( Rt->LeftChild, x ); return Rt; }
		return Rt;
	}
	int QueryR( node *Rt, member x ) {
		int Ans = 0;
		for( ; Rt != NULL; ) {
			if( Rt->Value == x ) return Ans + ( ( Rt->LeftChild != NULL ) ? Rt->LeftChild->Size : 0 ) + 1;
			if( Rt->Value < x ) {
				Ans += ( ( Rt->LeftChild != NULL ) ? Rt->LeftChild->Size : 0 ) + Rt->Cnt;
				Rt = Rt->RightChild;
			} else Rt = Rt->LeftChild;
		}
		return Ans + 1;
	}
	member QueryN( node *Rt, int x ) {
		for( ; Rt != NULL; ) {
			int Rc = 0; if( Rt->LeftChild != NULL ) Rc = Rt->LeftChild->Size;
			if( x > Rc && x <= Rc + Rt->Cnt ) return Rt->Value;
			if( x <= Rc ) Rt = Rt->LeftChild; else { x -= Rc + Rt->Cnt; Rt = Rt->RightChild; }
		}
		printf( "QueryNumber Failed.
" );
		return ( member ){ -1, -1 };
	}
	member pre( node *Rt, member x ) {
		member Ans = x;
		for( ; Rt != NULL; ) if( Rt->Value < x ) { Ans = Rt->Value; Rt = Rt->RightChild; } else Rt = Rt -> LeftChild;
		if( Ans == x ) printf( "Query Pre Failed.
" );
		return Ans;
	}
	member suc( node *Rt, member x ) {
		member Ans = x;
		for( ; Rt != NULL; ) if( Rt->Value > x ) { Ans = Rt->Value; Rt = Rt->LeftChild; } else Rt = Rt -> RightChild;
		if( Ans == x ) printf( "Query Suc Failed.
" );
		return Ans;
	}
	struct treap {
		node *Root;
		void clear() { delete [] Root; Root = NULL; srand( time( NULL ) ); return; }
		void insert( member x ) { Root = Insert( Root, x ); return; }
		void Delete( member x ) { Root = Del( Root, x ); return; }
		int QueryRank( member x ) { return QueryR( Root, x ); }
		member QueryNumber( int x ) { return QueryN( Root, x ); }
		member Pre( member x ) { return pre( Root, x ); }
		member Suc( member x ) { return suc( Root, x ); }
	};
} //Treap

Treap::treap Tree;

namespace UI {
	typedef unsigned int ui ;
	ui randNum( ui& seed , ui last , const ui m){ 
    	seed = seed * 17 + last ; return seed % m + 1; 
	}
	ui seed, last = 7;
	void InSeed() { scanf( "%llu", &seed ); return; }
} //UI

const int Maxm = 100010;
Treap::member Rec[ Maxm ];

void MAIN() {
	Tree.clear();
	int n, m; scanf( "%d%d", &m, &n ); UI::InSeed();
	for( int i = 1; i <= m; ++i ) {
		Tree.insert( ( Treap::member ){ 0, 0 } );
		Rec[ i ] = ( Treap::member ){ 0, 0 };
	}
	for( int i = 1; i <= n; ++i ) {
		int x = UI::randNum( UI::seed, UI::last, m );
		int y = UI::randNum( UI::seed, UI::last, m );
		Tree.Delete( Rec[ x ] );
		++Rec[ x ].Number;
		Rec[ x ].Time += y;
		Tree.insert( Rec[ x ] );
		UI::last = Tree.QueryRank( Rec[ x ] ) - 1;
		printf( "%llu
", UI::last );
	}
	return;
}

int main() {
	int TestCases; scanf( "%d", &TestCases );
	for( ; TestCases--; ) MAIN();
	return 0;
}
原文地址:https://www.cnblogs.com/chy-2003/p/11215514.html