BZOJ 2631: tree( LCT )

LCT...略麻烦...

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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 ) )

  

using namespace std;

 

const int maxn = 100000 + 5;

const int maxnode = maxn + 100;

const int MOD = 51061;

const int maxedge = maxn << 1;

 

#define mod( x ) ( ( x ) %= MOD )

 

int n;

 

struct edge {

int to;

edge* next;

};

 

edge* head[ maxn ];

edge EDGE[ maxedge ];

edge* e_pt;

 

void add( int u , int v ) {

e_pt -> to = v;

e_pt -> next = head[ u ];

head[ u ] = e_pt++;

}

 

#define add_edge( u , v ) add( u , v ) , add( v , u )

 

void edge_init() {

e_pt = EDGE;

clr( head , 0 );

}

 

struct Node *pt , *null;

 

struct Node {

Node *ch[ 2 ] , *p , *fa;

int add , mul , sum , v , s;

bool isRoot , rev;

Node() : v( 1 ) , add( 0 ) , mul( 1 ) , sum( 1 ) , s( 1 ) {

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

isRoot = true;

rev = false;

}

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

ch[ d ] = c;

c -> p = this;

}

inline bool d() {

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

}

inline void relax() {

if( rev ) {

rev = false;

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

   ch[ i ] -> Rev();

}

if( mul != 1 ) {

rep( i , 2 ) if( ch[ i ] != null ) 

   ch[ i ] -> Mul( mul );

mul = 1;

}

if( add ) {

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

   ch[ i ] -> Add( add );

add = 0;

}

}

inline void upd() {

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

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

}

inline void Rev() {

rev ^= 1;

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

}

inline void Add( int ad ) {

mod( add += ad );

mod( sum += 1LL * ad * s % MOD );

mod( v += ad );

}

inline void Mul( int mu ) {

add = 1LL * add * mu % MOD;

mul = 1LL * mul * mu % MOD;

sum = 1LL * sum * mu % MOD;

v = 1LL * v * mu % MOD;

}

inline void setRoot() {

fa = p;

p = null;

isRoot = true;

}

void* operator new( size_t ) {

return pt++;

}

};

 

Node NODE[ maxnode ];

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 ^ 1 ] , d );

t -> setc( p , d ^ 1 );

p -> upd();

if( p -> isRoot ) {

p -> isRoot = false;

t -> isRoot = true;

t -> fa = p -> fa;

}

}

 

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

static Node* S[ maxn ];

int top = 0;

for( Node* o = t ; o != null ; o = o -> p ) S[ ++top ] = o;

while( top ) S[ top-- ] -> relax();

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();

}

 

void access( Node* t ) {

for( Node* o = null ; t != null ; o = t , t = t -> fa ) {

splay( t );

t -> ch[ 1 ] -> setRoot();

t -> setc( o , 1 );

}

}

 

void makeRoot( Node* t ) {

access( t );

splay( t );

t -> Rev();

}

 

Node* findRoot( Node* t ) {

access( t );

splay( t );

for( ; t -> ch[ 0 ] != null ; t = t -> ch[ 0 ] )

   t -> relax();

splay( t );

return t;

}

 

void cut( Node* x , Node* y ) {

makeRoot( x );

access( y );

splay( x );

x -> setc( null , 1 );

x -> upd();

y -> p = null;

y -> setRoot();

}

 

void join( Node* x , Node* y ) {

makeRoot( x );

x -> fa = y;

}

 

Node* get( Node* x , Node* y ) {

makeRoot( x );

access( y );

splay( y );

return y;

}

 

void node_init() {

pt = NODE;

null = new( Node );

null -> s = null -> v = null -> sum = 0;

}

 

Node* V[ maxn ];

 

void dfs( int x , int par ) {

for( edge* e = head[ x ] ; e ; e = e -> next ) {

int to = e -> to;

if( to == par ) continue;

dfs( to , x );

V[ to ] -> fa = V[ x ];

}

}

 

void build_tree() {

rep( i , n ) V[ i ] = new( Node );

dfs( 0 , -1 );

}

 

int main() {

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

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

edge_init();

node_init();

int q , u , v;

cin >> n >> q;

rep( i , n - 1 ) {

scanf( "%d%d" , &u , &v );

u-- , v--;

add_edge( u , v );

}

build_tree();

char op;

while( q-- ) {

scanf( " %c%d%d" , &op , &u , &v );

u-- , v--;

if( op == '-' ) {

cut( V[ u ] , V[ v ] );

scanf( "%d%d" , &u , &v );

u-- , v--;

join( V[ u ] , V[ v ] );

} else if( op == '/' ) {

printf( "%d " , get( V[ u ] , V[ v ] ) -> sum );

} else {

int c;

scanf( "%d" , &c );

mod( c );

op != '+' ? get( V[ u ] , V[ v ] ) -> Mul( c ) : get( V[ u ] , V[ v ] ) -> Add( c );

}

}

return 0;

}

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2631: tree

Time Limit: 30 Sec  Memory Limit: 128 MB
Submit: 2454  Solved: 819
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Description

　一棵n个点的树，每个点的初始权值为1。对于这棵树有q个操作，每个操作为以下四种操作之一：
+ u v c：将u到v的路径上的点的权值都加上自然数c；
- u1 v1 u2 v2：将树中原有的边(u1,v1)删除，加入一条新边(u2,v2)，保证操作完之后仍然是一棵树；
* u v c：将u到v的路径上的点的权值都乘上自然数c；
/ u v：询问u到v的路径上的点的权值和，求出答案对于51061的余数。

Input

　　第一行两个整数n，q
接下来n-1行每行两个正整数u，v，描述这棵树
接下来q行，每行描述一个操作

Output

　　对于每个/对应的答案输出一行

Sample Input

3 2
1 2
2 3
* 1 3 4
/ 1 1

Sample Output

4

HINT

数据规模和约定

10%的数据保证，1<=n，q<=2000

另外15%的数据保证，1<=n，q<=5*10^4，没有-操作，并且初始树为一条链

另外35%的数据保证，1<=n，q<=5*10^4，没有-操作

100%的数据保证，1<=n，q<=10^5，0<=c<=10^4

Source