【HDU】5126 stars cdq分治套cdq分治套树状数组
传送门：【HDU】5126 stars
题目分析：将一个立方体查询拆成八个查询，然后对所有的操作+询问进行cdq分治。
由于是三维的，所以用cdq套cdq解决两维，最后一维用树状数组维护。
具体写法看代码好了，简单易懂。
cdq分治就是神，每套一层cdq分治，就可以将一维变成一个log。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long LL ;
#define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define ls ( o << 1 )
#define rs ( o << 1 | 1 )
#define lson ls , l , m
#define rson rs , m + 1 , r
#define root 1 , 1 , cnt
#define mid ( ( l + r ) >> 1 )
const int MAXN = 50005 ;
struct Node {
int x , y , z , f , idx ;
Node () {}
Node ( int x , int y , int z , int f , int idx ) : x ( x ) , y ( y ) , z ( z ) , f ( f ) , idx ( idx ) {}
} ;
Node q[MAXN * 8] , s1[MAXN * 8] , s2[MAXN * 8] , s[MAXN * 8] ;
int top1 , top2 , top ;
int m ;
int a[MAXN << 1] , cnt ;
int T[MAXN << 1] ;
int ans[MAXN] ;
int n ;
int unique ( int n ) {
int cnt = 1 ;
sort ( a + 1 , a + n + 1 ) ;
For ( i , 2 , n ) if ( a[i] != a[cnt] ) a[++ cnt] = a[i] ;
return cnt ;
}
int hash ( int x , int l = 1 , int r = cnt ) {
while ( l < r ) {
int m = mid ;
if ( a[m] >= x ) r = m ;
else l = m + 1 ;
}
return l ;
}
bool cmpz ( const Node& a , const Node& b ) {
if( a.z != b.z ) return a.z < b.z ;
return a.idx < b.idx ;
}
bool cmpy ( const Node& a , const Node& b ) {
if ( a.y != b.y ) return a.y < b.y ;
return a.idx < b.idx ;
}
void add ( int x , int v ) {
for ( int i = x ; i <= cnt ; i += i & -i ) T[i] += v ;
}
int sum ( int x , int ans = 0 ) {
for ( int i = x ; i > 0 ; i -= i & -i ) ans += T[i] ;
return ans ;
}
void cdq_fz2 ( int l , int r ) {
if ( r <= l ) return ;
int m = mid ;
cdq_fz2 ( l , m ) ;
cdq_fz2 ( m + 1 , r ) ;
top1 = top2 = 0 ;
For ( i , l , m ) if ( !s[i].f ) s1[top1 ++] = s[i] ;
For ( i , m + 1 , r ) if ( s[i].f ) s2[top2 ++] = s[i] ;
sort ( s1 , s1 + top1 , cmpy ) ;
sort ( s2 , s2 + top2 , cmpy ) ;
int j = 0 ;
rep ( i , 0 , top2 ) {
while ( j < top1 && s1[j].y <= s2[i].y ) add ( s1[j ++].x , 1 ) ;
ans[s2[i].idx] += s2[i].f * sum ( s2[i].x ) ;
}
rep ( i , 0 , j ) add ( s1[i].x , -1 ) ;;
}
void cdq_fz ( int l , int r ) {
if ( r <= l ) return ;
int m = mid ;
cdq_fz ( l , m ) ;
cdq_fz ( m + 1 , r ) ;
int top = 0 ;
For ( i , l , m ) if ( !q[i].f ) s[top ++] = q[i] ;
For ( i , m + 1 , r ) if ( q[i].f ) s[top ++] = q[i] ;
sort ( s , s + top , cmpz ) ;
cdq_fz2 ( 0 , top - 1 ) ;
}
void solve () {
int op ;
int x1 , y1 , z1 ;
int x2 , y2 , z2 ;
clr ( T , 0 ) ;
clr ( ans , 0 ) ;
cnt = 0 ;
m = 0 ;
scanf ( "%d" , &n ) ;
For ( i , 1 , n ) {
scanf ( "%d" , &op ) ;
if ( op == 1 ) {
scanf ( "%d%d%d" , &x1 , &y1 , &z1 ) ;
a[++ cnt] = x1 ;
q[++ m] = Node ( x1 , y1 , z1 , 0 , 0 ) ;
ans[i] = -1 ;
} else {
scanf ( "%d%d%d%d%d%d" , &x1 , &y1 , &z1 , &x2 , &y2 , &z2 ) ;
q[++ m] = Node ( x2 , y2 , z2 , 1 , i ) ;
q[++ m] = Node ( x2 , y2 , z1 - 1 , -1 , i ) ;
q[++ m] = Node ( x1 - 1 , y2 , z2 , -1 , i ) ;
q[++ m] = Node ( x2 , y1 - 1 , z2 , -1 , i ) ;
q[++ m] = Node ( x2 , y1 - 1 , z1 - 1 , 1 , i ) ;
q[++ m] = Node ( x1 - 1 , y2 , z1 - 1 , 1 , i ) ;
q[++ m] = Node ( x1 - 1 , y1 - 1 , z2 , 1 , i ) ;
q[++ m] = Node ( x1 - 1 , y1 - 1 , z1 - 1 , -1 , i ) ;
a[++ cnt] = x2 ;
a[++ cnt] = x1 - 1 ;
}
}
cnt = unique ( cnt ) ;
For ( i , 1 , m ) q[i].x = hash ( q[i].x ) ;
cdq_fz ( 1 , m ) ;
For ( i , 1 , n ) if ( ~ans[i] ) printf ( "%d\n" , ans[i] ) ;
}
int main () {
int T ;
scanf ( "%d" , &T ) ;
while ( T -- ) solve () ;
return 0 ;
}
--------------------------------update--------------------------------
给一个CDQ套BIT套Treap的代码吧。常数小的不得了。
代码如下：
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long LL ;
#define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define rev( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define lson l , m
#define rson m + 1 , r
#define mid ( ( l + r ) >> 1 )
const int MAXN = 50005 ;
struct Query {
int x1 , y1 ;
int x2 , y2 ;
int z , f , idx ;
Query () {}
Query ( int x1 , int y1 , int x2 , int y2 , int z , int f , int idx ) :
x1 ( x1 ) , y1 ( y1 ) , x2 ( x2 ) , y2 ( y2 ) , z ( z ) , f ( f ) , idx ( idx ) {}
} ;
struct Node* null ;
struct Node {
Node* c[2] ;
int r ;//greater root
int sum ;
int v ;
int key ;
void newnode ( int x , int value ) {
r = rand () ;
key = x ;
sum = v = value ;
c[0] = c[1] = null ;
}
void push_up () {
sum = c[0]->sum + v + c[1]->sum ;
}
} ;
Query opp[MAXN * 2] , s1[MAXN * 2] , s2[MAXN * 2] ;
Node pool[MAXN * 20] ;
Node* cur ;
Node* T[MAXN << 1] ;
int vis[MAXN << 1] , Time ;
int a[MAXN << 1] , cnt ;
int ans[MAXN] ;
int n , m , q ;
void rot ( Node* &o , int d ) {
Node* ch = o->c[d ^ 1] ;
o->c[d ^ 1] = ch->c[d] ;
ch->c[d] = o ;
o->push_up () ;
ch->push_up () ;
o = ch ;
}
void insert ( Node* &o , int x , int v ) {
if ( o == null ) {
o = cur ++ ;
o->newnode ( x , v ) ;
} else if ( o->key == x ) {
o->v += v ;
} else {
int d = ( o->key < x ) ;
insert ( o->c[d] , x , v ) ;
if ( o->c[d]->r > o->r ) rot ( o , d ^ 1 ) ;
}
o->push_up () ;
}
int search ( Node* o , int x , int ans = 0 ) {
while ( o != null ) {
if ( x < o->key ) {
o = o->c[0] ;
} else {
ans += o->c[0]->sum + o->v ;
o = o->c[1] ;
}
}
return ans ;
}
int unique ( int n ) {
int cnt = 1 ;
sort ( a + 1 , a + n + 1 ) ;
For ( i , 2 , n ) if ( a[i] != a[cnt] ) a[++ cnt] = a[i] ;
return cnt ;
}
int hash ( int x , int l = 1 , int r = cnt ) {
while ( l < r ) {
int m = mid ;
if ( a[m] >= x ) r = m ;
else l = m + 1 ;
}
return l ;
}
int cmpz ( const Query& a , const Query& b ) {
if ( a.z != b.z ) return a.z < b.z ;
return a.idx < b.idx ;
}
void add ( int x , int y , int v ) {
for ( int i = x ; i <= cnt ; i += i & -i ) {
if ( vis[i] != Time ) {
T[i] = null ;
vis[i] = Time ;
}
insert ( T[i] , y , v ) ;
}
}
int sum ( int x , int y , int ans = 0 ) {
for ( int i = x ; i > 0 ; i -= i & -i ) if ( vis[i] == Time ) ans += search ( T[i] , y ) ;
return ans ;
}
void cdq_fz ( int l , int r ) {
if ( r <= l ) return ;
int m = mid , top1 = 0 , top2 = 0 , j = 0 ;
cdq_fz ( lson ) ;
cdq_fz ( rson ) ;
For ( i , m + 1 , r ) if ( opp[i].idx ) s1[top1 ++] = opp[i] ;
For ( i , l , m ) if ( !opp[i].idx ) s2[top2 ++] = opp[i] ;
sort ( s1 , s1 + top1 , cmpz ) ;
sort ( s2 , s2 + top2 , cmpz ) ;
++ Time ;
cur = pool + 1 ;
rep ( i , 0 , top1 ) {
while ( j < top2 && s2[j].z <= s1[i].z ) {
add ( s2[j].x2 , s2[j].y2 , s2[j].f ) ;
++ j ;
}
ans[s1[i].idx] += s1[i].f * sum ( s1[i].x2 , s1[i].y2 ) ;
ans[s1[i].idx] -= s1[i].f * sum ( s1[i].x1 - 1 , s1[i].y2 ) ;
ans[s1[i].idx] -= s1[i].f * sum ( s1[i].x2 , s1[i].y1 - 1 ) ;
ans[s1[i].idx] += s1[i].f * sum ( s1[i].x1 - 1 , s1[i].y1 - 1 ) ;
}
}
void init () {
m = 0 ;
cnt = 0 ;
//srand ( 233333333 ) ;
cur = pool ;
null = cur ++ ;
null->c[0] = null->c[1] = null ;
null->sum = null->v = null->r = 0 ;
}
void scanf ( int& x , char c = 0 ) {
while ( ( c = getchar () ) < '0' ) ;
x = c - '0' ;
while ( ( c = getchar () ) >= '0' ) x = x * 10 + c - '0' ;
}
void solve () {
int op , x1 , y1 , z1 , x2 , y2 , z2 ;
init () ;
scanf ( n ) ;
For ( i , 1 , n ) {
scanf ( "%d" , &op ) ;
if ( op == 1 ) {
scanf ( x1 ) ;
scanf ( y1 ) ;
scanf ( z1 ) ;
opp[++ m] = Query ( x1 , y1 , x1 , y1 , z1 , 1 , 0 ) ;
a[++ cnt] = x1 ;
ans[i] = -1 ;
} else {
scanf ( x1 ) ;
scanf ( y1 ) ;
scanf ( z1 ) ;
scanf ( x2 ) ;
scanf ( y2 ) ;
scanf ( z2 ) ;
opp[++ m] = Query ( x1 , y1 , x2 , y2 , z2 , 1 , i ) ;
opp[++ m] = Query ( x1 , y1 , x2 , y2 , z1 - 1 , -1 , i ) ;
a[++ cnt] = x2 ;
a[++ cnt] = x1 - 1 ;
ans[i] = 0 ;
}
}
cnt = unique ( cnt ) ;
For ( i , 1 , m ) {
if ( !opp[i].idx ) opp[i].x1 = opp[i].x2 = hash ( opp[i].x1 ) ;
else {
opp[i].x1 = hash ( opp[i].x1 - 1 ) + 1 ;
opp[i].x2 = hash ( opp[i].x2 ) ;
}
}
cdq_fz ( 1 , m ) ;
For ( i , 1 , n ) if ( ~ans[i] ) printf ( "%d\n" , ans[i] ) ;
}
int main () {
int T ;
Time = 0 ;
clr ( vis , 0 ) ;
scanf ( T ) ;
while ( T -- ) solve () ;
return 0 ;
}