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
[Submit][Status][Discuss]
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
3 4 5 1 6 2
Sample Output
4 6 4 5 6 6