BZOJ 1552: [Cerc2007]robotic sort( splay )

kpm大神说可以用块状链表写...但是我不会...写了个splay....

先离散化 , 然后splay结点加个min维护最小值 , 就可以了... 

( ps BZOJ 3506 题意一样 , 双倍经验 ） 

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#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

 

#define rep( i , n ) for( int i = 0 ; i < n ; ++i )

#define clr( x , c ) memset( x , c , sizeof( x ) )

#define Rep( i , n ) for( int i = 1 ; i <= n ; ++i )

 

using namespace std;

 

const int maxn = 100000 + 5;

const int maxnode = 101000;

const int inf = 0x7fffffff;

 

int n;

 

struct data {

int v , pos;

bool operator < ( const data &rhs ) const {

return ( v < rhs.v ) || ( v == rhs.v && pos < rhs.pos );

}

};

 

bool cmp( const data &a , const data &b ) {

return a.pos < b.pos;

}

 

data A[ maxn ];

 

struct Node *pt , *null;

 

struct Node {

Node *ch[ 2 ] , *p;

int mn , val , size;

bool rev;

Node( int v = inf ) {

mn = val = v;

ch[ 0 ] = ch[ 1 ] = null;

rev = 0;

}

inline void setc( Node* c , int d ) {

ch[ d ] = c;

c -> p = this;

}

inline bool d() {

return this == p -> ch[ 1 ];

}

inline void Rev() {

rev ^= 1;

}

inline void upd() {

size = ch[ 0 ] -> size + ch[ 1 ] -> size + 1;

mn = min( val , min( ch[ 0 ] -> mn , ch[ 1 ] -> mn ) );

}

inline void relax() {

if( rev ) {

swap( ch[ 0 ] , ch[ 1 ] );

rep( i , 2 ) if( ch[ i ] != null )

   ch[ i ] -> Rev();

rev = 0;

}

}

void* operator new( size_t ) {

return pt++;

}

};

 

Node NODE[ maxnode ];

Node* root;

 

Node* build( int l , int r ) {

if( l >= r )

   return null;

int m = ( l + r ) >> 1;

Node* t = new Node( A[ m ].v );

t -> setc( build( l , m ) , 0 );

t -> setc( build( m + 1 , r ) , 1 );

t -> upd();

return t;

}

 

void rot( Node* t ) {

Node* p = t -> p;

p -> relax();

t -> relax();

int d = t -> d();

p -> p -> setc( t , p -> d() );

p -> setc( t -> ch[ ! d ] , d );

t -> setc( p , ! d );

p -> upd();

if( p == root ) root = t;

}

 

void splay( Node* t , Node* f = null ) {

while( t -> p != f ) {

if( t -> p -> p == f ) rot( t );

else if( t -> d() != t -> p -> d() ) rot( t ) , rot( t );

else rot( t -> p ) , rot( t );

}

t -> upd();

}

 

Node* select( int k ) {

for( Node* t = root ; ; ) {

t -> relax();

int s = t -> ch[ 0 ] -> size;

if( s == k ) return t;

else if( s < k ) {

k -= s + 1;

t = t -> ch[ 1 ];

} else 

   t = t -> ch[ 0 ];

}

}

 

int v;

Node* find( Node* t ) {

t -> relax();

if( t -> val == v ) return t;

else return find( t -> ch[ 0 ] -> mn != v ? t -> ch[ 1 ] : t -> ch[ 0 ] );

}

 

Node* &get( int l , int r ) {

l-- , r++;

Node* L = select( l );

Node* R = select( r );

splay( L );

splay( R , L );

return R -> ch[ 0 ];

}

 

void init() {

pt = NODE;

null = new( Node );

null -> size = 0;

root = build( 0 , n + 2 );

root -> p = null;

}

 

int main() {

freopen( "test.in" , "r" , stdin );

cin >> n;

Rep( i , n ) {

scanf( "%d" , &A[ i ].v );

A[ i ].pos = i;

}

sort( A + 1 , A + n + 1 );

Rep( i , n ) A[ i ].v = i;

sort( A + 1 , A + n + 1 , cmp );

A[ 0 ].v = A[ n + 1 ].v = inf;

init();

Rep( i , n ) {

v = i;

Node* &t = get( i , n );

Node* x = find( t );

splay( x );

int k = x -> ch[ 0 ] -> size;

printf( "%d" , k );

if( i != n ) printf( " " );

x = get( i , k );

x -> Rev();

splay( x );

}

return 0;

}

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1552: [Cerc2007]robotic sort

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 486  Solved: 203
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Description

Input

输入共两行，第一行为一个整数N，N表示物品的个数，1<=N<=100000。第二行为N个用空格隔开的正整数，表示N个物品最初排列的编号。

Output

输出共一行，N个用空格隔开的正整数P1,P2,P3…Pn，（1 < = Pi < = N），Pi表示第i次操作前第i小的物品所在的位置。 注意：如果第i次操作前，第i小的物品己经在正确的位置Pi上，我们将区间[Pi,Pi]反转(单个物品)。

Sample Input

6
3 4 5 1 6 2

Sample Output

4 6 4 5 6 6

HINT

Source

HNOI2009集训Day6