BZOJ 1500: [NOI2005]维修数列( splay )

splay..... 

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

#include<queue>

 

#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 = 500000;

const int maxnode = 550000;

const int inf = int( 1e7 );

 

struct Node* null;

queue< Node* > Q;

int seq[ maxn ] , tot , pos;

 

struct Node {

Node *ch[ 2 ] , *p;

int s , v , sum , mx , l_m , r_m;

bool rev , tag;

Node( int _v = 0 ) {

l_m = r_m = sum = mx = v = _v;

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

rev = tag = false;

}

inline void relax() {

if( rev ) {

rev = false;

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

   ch[ i ] -> Rev();

}

if( tag ) {

tag = false;

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

   ch[ i ] -> Set( v );

}

}

inline void Rev() {

rev ^= 1;

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

swap( l_m , r_m );

}

inline void Set( int _v ) {

tag = true;

v = _v;

sum = s * _v;

mx = r_m = l_m = _v * ( _v > 0 ? s : 1 );

}

inline void upd() {

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

sum = ch[ 0 ] -> sum + ch[ 1 ] -> sum + v;

l_m = max( ch[ 0 ] -> l_m , ch[ 0 ] -> sum + v + max( 0 , ch[ 1 ] -> l_m ) );

r_m = max( ch[ 1 ] -> r_m , ch[ 1 ] -> sum + v + max( 0 , ch[ 0 ] -> r_m ) );

mx = max( 0 , ch[ 0 ] -> r_m ) + v + max( 0 , ch[ 1 ] -> l_m );

mx = max( mx , max( ch[ 0 ] -> mx , ch[ 1 ] -> mx ) );

}

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

ch[ d ] = c;

c -> p = this;

}

inline bool d() {

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

}

void* operator new( size_t ) {

Node* t = Q.front();

Q.pop();

return t;

}

};

 

Node memory[ maxnode ];

Node* root;

 

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 ) {

for( Node* p = t -> p ; p != f ; p = t -> p ) {

if( p -> p != f )

   p -> d() != t -> d() ? rot( t ) : rot( p );

rot( t );

}

t -> upd();

}

 

Node* select( int k ) {

for( Node* t = root ; ; ) {

t -> relax();

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

if( k == s ) return t;

if( k > s )

   k -= s + 1 , t = t ->ch[ 1 ];

else 

   t = t -> ch[ 0 ];

}

}

 

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

l-- , r++;

Node *L = select( l ) , *R = select( r );

splay( L );

splay( R , L );

return R -> ch[ 0 ];

}

 

Node* build( int l , int r ) {

if( l >= r ) return null;

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

Node* t = new Node( seq[ m ] );

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

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

t -> upd();

return t;

}

 

void ins() {

Node *L = select( pos ) , *R = select( pos + 1 );

splay( L ) , splay( R , L );

R -> setc( build( 1 , tot + 1 ) , 0 );

R -> upd();

L -> upd();

}

 

void del( Node* t ) {

if( t == null ) return;

rep( i , 2 ) del( t -> ch[ i ] );

Q.push( t );

}

 

void init() {

rep( i , maxnode ) Q.push( memory + i );

null = new Node( -inf );

null -> s = 0;

null -> sum = 0;

}

 

int main() {

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

// freopen( "test.out" , "w" , stdout );

init();

int n , m;

cin >> n >> m;

Rep( i , n ) scanf( "%d" , seq + i );

root = build( 0 , n + 2 );

char S[ 15 ];

while( m-- ) {

scanf( " %s" , S );

if( S[ 2 ] == 'X' ) {

Node* &t = get( 1 , n );

printf( "%d " , t -> mx );

} else {

scanf( "%d%d" , &pos , &tot );

if( S[ 0 ] == 'I' ) {

n += tot;

Rep( i , tot ) scanf( "%d" , seq + i );

ins();

} else if( S[ 0 ] == 'D' ) {

Node* &t = get( pos , pos + tot - 1 );

n -= tot;

del( t );

root -> ch[ 1 ] -> setc( null , 0 );

splay( root -> ch[ 1 ] );

} else if( S[ 0 ] == 'R' ) {

Node* &t = get( pos , pos + tot - 1 );

t -> Rev();

splay( t );

} else if( S[ 0 ] == 'M' ) {

Node* &t = get( pos , pos + tot - 1 );

int v;

scanf( "%d" , &v );

t -> Set( v );

splay( t );

} else if( S[ 0 ] == 'G' ) {

Node* &t = get( pos , pos + tot -1 );

printf( "%d " , t -> sum );

}

}

}

return 0;

}

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1500: [NOI2005]维修数列

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 8501  Solved: 2563
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Description

Input

输入文件的第1行包含两个数N和M，N表示初始时数列中数的个数，M表示要进行的操作数目。第2行包含N个数字，描述初始时的数列。以下M行，每行一条命令，格式参见问题描述中的表格。

Output

对于输入数据中的GET-SUM和MAX-SUM操作，向输出文件依次打印结果，每个答案（数字）占一行。

Sample Input

9 8
2 -6 3 5 1 -5 -3 6 3
GET-SUM 5 4
MAX-SUM
INSERT 8 3 -5 7 2
DELETE 12 1
MAKE-SAME 3 3 2
REVERSE 3 6
GET-SUM 5 4
MAX-SUM

Sample Output

-1
10
1
10

HINT

Source

Splay