「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\(O(n\log n)\)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\n", 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.\n" );
        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.\n" );
        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.\n" );
        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\n", 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