NOI2004郁闷的出纳员

传送门
题目看起来玄乎,但其实只需要一点点小 trick 就可以了.

我们可以用一个全局的 delta 来维护工资的调整记录

对于每一个新加入的员工,先判断是否低于最低工资下限,如果是,直接踢出,不做任何操作,否则,将其插入 Treap 中,不过这时为了不对以后的查询产生影响,我们要插入的值时 key-delta (想一想,为什么?)

对于加工资的操作,直接在 delta 上统计即可.而减工资,这可能牵扯到会有员工离开,所以我们不能只修改delta

我们在修改 delta 之后,把 Treap 按权值分割,分割标准是 minn-delta-1 (想一想,为什么?提示:不等式移项)

然后直接舍弃整个的左子树,此时的左子树就是所有会离开公司的员工代表的节点,所以最后的答案要加上该子树的 size

对于查询操作,直接查询出来的第 k 小(注意!!!是第 k 小)加上 delta 即可

Code:

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <ctime>
#define Drt pair < Treap * , Treap * >
#define siz(rt) ( rt == NULL ? 0 : rt->size )
#define int long long

using std :: pair ;

int n , minn , cnt , delta ;

struct Treap {
    Treap * son[2] ;
    int val , size , rank ;
    Treap (int val) : val ( val ) { size = 1 ; son[0] = son[1] = NULL ; rank = rand () ; }
    inline void maintain () {
        this->size = 1 ;
        if ( this->son[0] != NULL ) this->size += this->son[0]->size ;
        if ( this->son[1] != NULL ) this->size += this->son[1]->size ;
        return ;
    }
} * root = NULL ;

inline Drt Split ( Treap * rt , int k ) {
    if ( rt == NULL ) return Drt ( NULL , NULL ) ;
    Drt t ;
    if ( k <= siz ( rt->son[0] ) ) {
        t = Split ( rt->son[0] , k ) ; rt->son[0] = t.second ;
        rt->maintain () ; t.second = rt ;
    } else {
        t = Split ( rt->son[1] , k - siz ( rt->son[0] ) - 1 ) ;
        rt->son[1] = t.first ; rt->maintain () ; t.first = rt ;
    }
    return t ;
}

inline Drt SplitV ( Treap * rt , int key ) {
    if ( rt == NULL ) return Drt ( NULL , NULL ) ;
    Drt t ;
    if ( rt->val <= key ) {
        t = SplitV ( rt->son[1] , key ) ; rt->son[1] = t.first ;
        rt->maintain () ; t.first = rt ;
    } else {
        t = SplitV ( rt->son[0] , key ) ; rt->son[0] = t.second ;
        rt->maintain () ; t.second = rt ;
    }
    return t ;
}

inline Treap * merge ( Treap * x , Treap * y ) {
    if ( x == NULL ) return y ; if ( y == NULL ) return x ;
    if ( x->rank < y->rank ) {
        x->son[1] = merge ( x->son[1] , y ) ;
        x->maintain () ; return x ;
    } else {
        y->son[0] = merge ( x , y->son[0] ) ;
        y->maintain () ; return y ;
    }
}

inline int Getrank ( Treap * rt , int key ) {
    if ( rt == NULL ) return 0 ;
    if ( key <= rt->val ) return Getrank ( rt->son[0] , key ) ;
    else return Getrank ( rt->son[1] , key ) + siz ( rt->son[0] ) + 1 ;
}

inline int Getkth ( Treap * & rt , int key ) {
    Drt x = Split ( rt , key - 1 ) ;
    Drt y = Split ( x.second , 1 ) ;
    Treap * node = y.first ;
    rt = merge ( x.first , merge ( node , y.second ) ) ; 
    return node == NULL ? 0 : node->val ;
}

inline void insert ( Treap * & rt , int key ) {
    int k = Getrank ( rt , key ) ; Drt t = Split ( rt , k ) ;
    Treap * node = new Treap ( key ) ;
    rt = merge ( t.first , merge ( node , t.second ) ) ;
    return ; 
}

signed main () {
    scanf ("%lld%lld" , & n , & minn ) ;
    while ( n -- ) {
        char opt[4] ; int key ;
        scanf ("%s%lld" , opt , & key ) ;
        if ( opt[0] == 'I') {
            if ( key < minn ) continue ;
            insert ( root , key - delta ) ;
        }
        if ( opt[0] == 'A') delta += key ;
        if ( opt[0] == 'S') {
            delta -= key ;
            Drt t = SplitV ( root , minn - delta - 1 ) ;
            root = t.second ; cnt += siz ( t.first )  ;
        }
        if ( opt[0] == 'F') {
            if ( key > siz ( root ) ) printf ("-1\n") ; 
            else printf ("%lld\n" , Getkth ( root , siz ( root ) - key + 1 ) + delta ) ;
        }
    }
    printf ("%lld\n" , cnt ) ;
    system ("pause") ; return 0 ;
}

标签：rt,return,郁闷,出纳员,son,Treap,NOI2004,key,NULL
来源： https://www.cnblogs.com/Equinox-Flower/p/10785300.html