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codeforces contest 1140（D~G）
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前言
A~C不想寫博客了，就不寫了，后面的題還是要推一推的，所以寫一下
CF 1140 D
題目大意 ：11 1 nn n n≤500n\le500 n ≤ 5 0 0 題解： O(n3)\mathcal O(n^3) O ( n 3 ) O(1)\mathcal O(1) O ( 1 ) 11 1 nn n Δ\Delta Δ x≠n?1x\neq n-1 x ? ? = n ? 1 Δ\Delta Δ k>xk>x k > x 此處貼了一幅圖，可以根據圖來理解下面的證明 ）Δ\Delta Δ Δ\Delta Δ Δ\Delta Δ Δ\Delta Δ 1?x?n+x?k?n>1?x?k+1?k?n1*x*n+x*k*n>1*x*k+1*k*n 1 ? x ? n + x ? k ? n > 1 ? x ? k + 1 ? k ? n x?k?n>1?k?nx*k*n>1*k*n x ? k ? n > 1 ? k ? n 1?x?n>1?x?k1*x*n>1*x*k 1 ? x ? n > 1 ? x ? k xx x n?1n-1 n ? 1 nn n 11 1 11 1 ∏i=2n?1i?(i+1)\prod_{i=2}^{n-1}i*(i+1) i = 2 ∏ n ? 1 ? i ? ( i + 1 ) ∏i=2n?1i2+i\prod_{i=2}^{n-1}i^2+i i = 2 ∏ n ? 1 ? i 2 + i
#include <bits/stdc++.h>
typedef long long ll
;
#define rg register
template
< typename T
> inline void read ( T
& x
) { char cu
= getchar ( ) ; x
= 0 ; bool fla
= 0 ; while ( ! isdigit ( cu
) ) { if ( cu
== '-' ) fla
= 1 ; cu
= getchar ( ) ; } while ( isdigit ( cu
) ) x
= x
* 10 + cu
- '0' , cu
= getchar ( ) ; if ( fla
) x
= - x
; }
template
< typename T
> inline void printe ( const T x
) { if ( x
>= 10 ) printe ( x
/ 10 ) ; putchar ( x
% 10 + '0' ) ; }
template
< typename T
> inline void print ( const T x
) { if ( x
< 0 ) putchar ( '-' ) , printe ( - x
) ; else printe ( x
) ; }
template
< typename T
> inline T
min ( const T a
, const T b
) { return a
< b
? a
: b
; }
template
< typename T
> inline T
max ( const T a
, const T b
) { return a
> b
? a
: b
; }
int n
;
int main ( )
{ read ( n
) ; print ( ( n
- 1 ) * n
* ( n
* 2 - 1 ) / 6 + n
* ( n
- 1 ) / 2 - 2 ) ; return 0 ;
}
CF 1140 E
題目大意 ：給定一個值域為kk k n,k≤2?105n,k\le2·10^5 n , k ≤ 2 ? 1 0 5 題解： 這里有個奇 長度回文子串，這是非常好的性質，我們發現如果對于任何ii i ai≠ai+2a_i\neq a_{i+2} a i ? ? ? = a i + 2 ? O(n)\mathcal O(n) O ( n )
#include <bits/stdc++.h>
typedef long long ll
;
#define rg register
template
< typename T
> inline void read ( T
& x
) { char cu
= getchar ( ) ; x
= 0 ; bool fla
= 0 ; while ( ! isdigit ( cu
) ) { if ( cu
== '-' ) fla
= 1 ; cu
= getchar ( ) ; } while ( isdigit ( cu
) ) x
= x
* 10 + cu
- '0' , cu
= getchar ( ) ; if ( fla
) x
= - x
; }
template
< typename T
> inline void printe ( const T x
) { if ( x
>= 10 ) printe ( x
/ 10 ) ; putchar ( x
% 10 + '0' ) ; }
template
< typename T
> inline void print ( const T x
) { if ( x
< 0 ) putchar ( '-' ) , printe ( - x
) ; else printe ( x
) ; }
template
< typename T
> inline T
min ( const T a
, const T b
) { return a
< b
? a
: b
; }
template
< typename T
> inline T
max ( const T a
, const T b
) { return a
> b
? a
: b
; }
const int mod
= 998244353 ;
int n
, k
, a
[ 200001 ] , b
[ 200001 ] , ta
, tb
;
ll A
[ 200001 ] , B
[ 200001 ] , C
[ 200001 ] ;
ll ans
= 1 ;
int main ( )
{ read ( n
) , read ( k
) ; for ( rg
int i
= 1 ; i
<= n
; i
++ ) if ( i
& 1 ) read ( a
[ ++ ta
] ) ; else read ( b
[ ++ tb
] ) ; A
[ 1 ] = k
- 1 , B
[ 1 ] = k
- 2 , C
[ 1 ] = k
- 1 ; A
[ 0 ] = B
[ 0 ] = C
[ 0 ] = 1 ; for ( rg
int i
= 2 ; i
<= n
; i
++ ) { A
[ i
] = B
[ i
- 1 ] * ( k
- 1 ) % mod
; B
[ i
] = ( B
[ i
- 1 ] * ( k
- 2 ) + A
[ i
- 1 ] ) % mod
; C
[ i
] = C
[ i
- 1 ] * ( k
- 1 ) % mod
; } for ( rg
int i
= 1 ; i
<= ta
; i
++ ) if ( a
[ i
] == - 1 ) { int j
= i
; while ( j
< ta
&& a
[ j
+ 1 ] == - 1 ) j
++ ; if ( i
== 1 && j
== ta
) ans
= ans
* C
[ ta
- 1 ] % mod
* k
% mod
; else if ( i
== 1 || j
== ta
) ans
= ans
* C
[ j
- i
+ 1 ] % mod
; else if ( a
[ i
- 1 ] == a
[ j
+ 1 ] ) ans
= ans
* A
[ j
- i
+ 1 ] % mod
; else ans
= ans
* B
[ j
- i
+ 1 ] % mod
; i
= j
+ 1 ; } else if ( i
!= ta
&& a
[ i
+ 1 ] != - 1 && a
[ i
] == a
[ i
+ 1 ] ) ans
= 0 ; for ( rg
int i
= 1 ; i
<= tb
; i
++ ) if ( b
[ i
] == - 1 ) { int j
= i
; while ( j
< tb
&& b
[ j
+ 1 ] == - 1 ) j
++ ; if ( i
== 1 && j
== tb
) ans
= ans
* C
[ tb
- 1 ] % mod
* k
% mod
; else if ( i
== 1 || j
== tb
) ans
= ans
* C
[ j
- i
+ 1 ] % mod
; else if ( b
[ i
- 1 ] == b
[ j
+ 1 ] ) ans
= ans
* A
[ j
- i
+ 1 ] % mod
; else ans
= ans
* B
[ j
- i
+ 1 ] % mod
; i
= j
+ 1 ; } else if ( i
!= tb
&& b
[ i
+ 1 ] != - 1 && b
[ i
] == b
[ i
+ 1 ] ) ans
= 0 ; print ( ans
) ; return 0 ;
}
CF 1140 F
題目大意 ：在平面直角坐標系中，如果有3個點能作為一個矩形（邊平行于坐標軸）的三個頂點，那么我們可以把這個矩形的第四個點也加入點集，直到不能再加入SS S TT T SS S ∣T∣|T| ∣ T ∣ q≤3?105，1≤xi,yi≤3?105q\le3·10^5，1\le x_i,y_i\le3·10^5 q ≤ 3 ? 1 0 5 ， 1 ≤ x i ? , y i ? ≤ 3 ? 1 0 5 題解： 我們考慮假如只有加點操作 考慮如何解決刪點的情況 O(logq)\mathcal O(logq) O ( l o g q ) O(logq)\mathcal O(logq) O ( l o g q ) O(qlog2q)\mathcal O(qlog^2q) O ( q l o g 2 q ) 本題后文有彩蛋，請一定要觀看 ）
#include <bits/stdc++.h>
typedef long long ll
;
#define rg register
template
< typename T
> inline void read ( T
& x
) { char cu
= getchar ( ) ; x
= 0 ; bool fla
= 0 ; while ( ! isdigit ( cu
) ) { if ( cu
== '-' ) fla
= 1 ; cu
= getchar ( ) ; } while ( isdigit ( cu
) ) x
= x
* 10 + cu
- '0' , cu
= getchar ( ) ; if ( fla
) x
= - x
; }
template
< typename T
> inline void printe ( const T x
) { if ( x
>= 10 ) printe ( x
/ 10 ) ; putchar ( x
% 10 + '0' ) ; }
template
< typename T
> inline void print ( const T x
) { if ( x
< 0 ) putchar ( '-' ) , printe ( - x
) ; else printe ( x
) ; }
template
< typename T
> inline T
min ( const T a
, const T b
) { return a
< b
? a
: b
; }
template
< typename T
> inline T
max ( const T a
, const T b
) { return a
> b
? a
: b
; }
const int MAX
= 2097152 ;
int q
, l
[ MAX
] , r
[ MAX
] , mid
[ MAX
] ;
std
: : vector
< std
: : pair
< int , int > > irt
[ MAX
] ;
std
: : vector
< std
: : pair
< int , int > > : : iterator pos
;
void ini ( const int root
, const int ll
, const int rr
)
{ l
[ root
] = ll
, r
[ root
] = rr
, mid
[ root
] = ( ll
+ rr
) >> 1 ; if ( ll
== rr
) return ; ini ( root
<< 1 , ll
, mid
[ root
] ) , ini ( root
<< 1 | 1 , mid
[ root
] + 1 , rr
) ;
}
void insert ( const int root
, const int wanl
, const int wanr
, std
: : pair
< int , int > V
)
{ if ( wanl
== l
[ root
] && wanr
== r
[ root
] ) { irt
[ root
] . push_back ( V
) ; return ; } if ( wanr
<= mid
[ root
] ) insert ( root
<< 1 , wanl
, wanr
, V
) ; else if ( wanl
> mid
[ root
] ) insert ( root
<< 1 | 1 , wanl
, wanr
, V
) ; else insert ( root
<< 1 , wanl
, mid
[ root
] , V
) , insert ( root
<< 1 | 1 , mid
[ root
] + 1 , wanr
, V
) ;
}
struct P
{ int a
, b
, t
; bool operator
< ( const P B
) const { return a
== B
. a
? b
< B
. b
: a
< B
. a
; }
} ;
std
: : set
< P
> Q
;
std
: : set
< P
> : : iterator Pos
;
struct info
{ int v
, a
, b
; void clear ( ) { v
= a
= b
= 0 ; }
} ;
struct node
{ node
* lson
, * rson
; info v
;
} t
[ 12600021 ] , * root
[ 600001 ] , empty
, * Null
;
int usdto
[ 600001 ] , id
;
node
* newnode ( )
{ node
* R
= & t
[ ++ usdto
[ id
] ] ; R
-> lson
= R
-> rson
= Null
, R
-> v
. clear ( ) ; return R
;
}
int pl
; info val
;
void INSERT ( node
* las
, node
* dq
, const int root
)
{ if ( l
[ root
] == r
[ root
] ) { dq
-> v
= val
; return ; } if ( pl
<= mid
[ root
] ) INSERT ( las
-> lson
, dq
-> lson
= newnode ( ) , root
<< 1 ) , dq
-> rson
= las
-> rson
; else INSERT ( las
-> rson
, dq
-> rson
= newnode ( ) , root
<< 1 | 1 ) , dq
-> lson
= las
-> lson
;
}
info RES
;
void SEARCH ( node
* dq
, const int root
, const int pl
)
{ if ( l
[ root
] == r
[ root
] ) { if ( dq
-> v
. v
== 0 ) RES
= ( info
) { pl
, pl
<= 300000 , pl
> 300000 } ; else RES
= dq
-> v
; return ; } if ( pl
<= mid
[ root
] ) SEARCH ( dq
-> lson
, root
<< 1 , pl
) ; else SEARCH ( dq
-> rson
, root
<< 1 | 1 , pl
) ;
}
ll ans
[ 600001 ] ;
info
find ( const int x
)
{ SEARCH ( root
[ id
] , 1 , x
) ; info u
= RES
; if ( u
. v
== x
) return u
; return find ( u
. v
) ;
}
void dodo ( std
: : pair
< int , int > V
)
{
V
. second
+ = 300000 ; info u
= find ( V
. first
) , v
= find ( V
. second
) ; if ( u
. v
== v
. v
) id
+ = 2 , usdto
[ id
] = usdto
[ id
- 2 ] , root
[ id
] = root
[ id
- 2 ] , ans
[ id
] = ans
[ id
- 2 ] ; else { if ( u
. a
+ u
. b
< v
. a
+ v
. b
) std
: : swap ( u
, v
) ; id
++ , usdto
[ id
] = usdto
[ id
- 1 ] ; root
[ id
] = newnode ( ) ; pl
= v
. v
, val
= ( info
) { u
. v
, 0 , 0 } ; INSERT ( root
[ id
- 1 ] , root
[ id
] , 1 ) ; id
++ , usdto
[ id
] = usdto
[ id
- 1 ] ; root
[ id
] = newnode ( ) ; pl
= u
. v
, val
= ( info
) { u
. v
, u
. a
+ v
. a
, u
. b
+ v
. b
} ; INSERT ( root
[ id
- 1 ] , root
[ id
] , 1 ) ; ans
[ id
] = ans
[ id
- 2 ] ; ans
[ id
] + = ( ll
) ( u
. a
+ v
. a
) * ( u
. b
+ v
. b
) - ( ll
) u
. a
* u
. b
- ( ll
) v
. a
* v
. b
; }
}
void undo ( )
{
id
- = 2 ;
}
void dfs ( const int root
)
{ if ( l
[ root
] > q
) return ;
for ( pos
= irt
[ root
] . begin ( ) ; pos
!= irt
[ root
] . end ( ) ; pos
++ ) dodo ( * pos
) ; if ( l
[ root
] == r
[ root
] ) print ( ans
[ id
] ) , putchar ( ' ' ) ; else dfs ( root
<< 1 ) , dfs ( root
<< 1 | 1 ) ; for ( pos
= irt
[ root
] . begin ( ) ; pos
!= irt
[ root
] . end ( ) ; pos
++ ) undo ( ) ;
}
int main ( )
{ empty
. lson
= empty
. rson
= Null
= & empty
; root
[ 0 ] = newnode ( ) ; read ( q
) ; ini ( 1 , 1 , 600000 ) ; for ( rg
int i
= 1 ; i
<= q
; i
++ ) { P u
; read ( u
. a
) , read ( u
. b
) ; if ( ( Pos
= Q
. find ( u
) ) == Q
. end ( ) ) { u
. t
= i
; Q
. insert ( u
) ; } else { insert ( 1 , ( * Pos
) . t
, i
- 1 , std
: : make_pair ( ( * Pos
) . a
, ( * Pos
) . b
) ) ; Q
. erase ( Pos
) ; } } while ( Q
. size ( ) ) { Pos
= Q
. begin ( ) ; insert ( 1 , ( * Pos
) . t
, q
, std
: : make_pair ( ( * Pos
) . a
, ( * Pos
) . b
) ) ; Q
. erase ( Pos
) ; } dfs ( 1 ) ; return 0 ;
}
xswl，我以為可持久化并查集是對的，其實是3個log的，T成大傻逼 上面貼出的是TLE代碼 O(1)\mathcal O(1) O ( 1 )
#include <bits/stdc++.h>
typedef long long ll
;
#define rg register
template
< typename T
> inline void read ( T
& x
) { char cu
= getchar ( ) ; x
= 0 ; bool fla
= 0 ; while ( ! isdigit ( cu
) ) { if ( cu
== '-' ) fla
= 1 ; cu
= getchar ( ) ; } while ( isdigit ( cu
) ) x
= x
* 10 + cu
- '0' , cu
= getchar ( ) ; if ( fla
) x
= - x
; }
template
< typename T
> inline void printe ( const T x
) { if ( x
>= 10 ) printe ( x
/ 10 ) ; putchar ( x
% 10 + '0' ) ; }
template
< typename T
> inline void print ( const T x
) { if ( x
< 0 ) putchar ( '-' ) , printe ( - x
) ; else printe ( x
) ; }
template
< typename T
> inline T
min ( const T a
, const T b
) { return a
< b
? a
: b
; }
template
< typename T
> inline T
max ( const T a
, const T b
) { return a
> b
? a
: b
; }
const int MAX
= 2097152 ;
int q
, l
[ MAX
] , r
[ MAX
] , mid
[ MAX
] ;
std
: : vector
< std
: : pair
< int , int > > irt
[ MAX
] ;
std
: : vector
< std
: : pair
< int , int > > : : iterator pos
;
void ini ( const int root
, const int ll
, const int rr
)
{ l
[ root
] = ll
, r
[ root
] = rr
, mid
[ root
] = ( ll
+ rr
) >> 1 ; if ( ll
== rr
) return ; ini ( root
<< 1 , ll
, mid
[ root
] ) , ini ( root
<< 1 | 1 , mid
[ root
] + 1 , rr
) ;
}
void insert ( const int root
, const int wanl
, const int wanr
, std
: : pair
< int , int > V
)
{ if ( wanl
== l
[ root
] && wanr
== r
[ root
] ) { irt
[ root
] . push_back ( V
) ; return ; } if ( wanr
<= mid
[ root
] ) insert ( root
<< 1 , wanl
, wanr
, V
) ; else if ( wanl
> mid
[ root
] ) insert ( root
<< 1 | 1 , wanl
, wanr
, V
) ; else insert ( root
<< 1 , wanl
, mid
[ root
] , V
) , insert ( root
<< 1 | 1 , mid
[ root
] + 1 , wanr
, V
) ;
}
struct P
{ int a
, b
, t
; bool operator
< ( const P B
) const { return a
== B
. a
? b
< B
. b
: a
< B
. a
; }
} ;
std
: : set
< P
> Q
;
std
: : set
< P
> : : iterator Pos
;
ll ans
;
int bcj
[ 600001 ] , ltot
[ 600001 ] , rtot
[ 600001 ] ;
int find ( const int x
)
{ if ( x
== bcj
[ x
] ) return x
; return find ( bcj
[ x
] ) ;
}
int cha
[ 600001 ] , chb
[ 600001 ] , opt
;
void dodo ( std
: : pair
< int , int > V
)
{ V
. second
+ = 300000 ; int u
= find ( V
. first
) , v
= find ( V
. second
) ; opt
++ ; if ( u
== v
) cha
[ opt
] = 0 ; else { if ( ltot
[ u
] + rtot
[ u
] < ltot
[ v
] + rtot
[ v
] ) std
: : swap ( u
, v
) ; bcj
[ v
] = u
; ans
+ = ( ll
) ( ltot
[ u
] + ltot
[ v
] ) * ( rtot
[ u
] + rtot
[ v
] ) - ( ll
) ltot
[ u
] * rtot
[ u
] - ( ll
) ltot
[ v
] * rtot
[ v
] ; ltot
[ u
] + = ltot
[ v
] ; rtot
[ u
] + = rtot
[ v
] ; cha
[ opt
] = u
, chb
[ opt
] = v
; }
}
void undo ( )
{ if ( cha
[ opt
] ) { int u
= cha
[ opt
] , v
= chb
[ opt
] ; ltot
[ u
] - = ltot
[ v
] ; rtot
[ u
] - = rtot
[ v
] ; ans
- = ( ll
) ( ltot
[ u
] + ltot
[ v
] ) * ( rtot
[ u
] + rtot
[ v
] ) - ( ll
) ltot
[ u
] * rtot
[ u
] - ( ll
) ltot
[ v
] * rtot
[ v
] ; bcj
[ v
] = v
; } opt
-- ;
}
void dfs ( const int root
)
{ for ( pos
= irt
[ root
] . begin ( ) ; pos
!= irt
[ root
] . end ( ) ; pos
++ ) dodo ( * pos
) ; if ( l
[ root
] == r
[ root
] ) print ( ans
) , putchar ( ' ' ) ; else dfs ( root
<< 1 ) , dfs ( root
<< 1 | 1 ) ; for ( pos
= irt
[ root
] . begin ( ) ; pos
!= irt
[ root
] . end ( ) ; pos
++ ) undo ( ) ;
}
int main ( )
{ for ( rg
int i
= 1 ; i
<= 600000 ; i
++ ) { bcj
[ i
] = i
; if ( i
<= 300000 ) ltot
[ i
] ++ ; else rtot
[ i
] ++ ; } read ( q
) ; ini ( 1 , 1 , q
) ; for ( rg
int i
= 1 ; i
<= q
; i
++ ) { P u
; read ( u
. a
) , read ( u
. b
) ; if ( ( Pos
= Q
. find ( u
) ) == Q
. end ( ) ) { u
. t
= i
; Q
. insert ( u
) ; } else { insert ( 1 , ( * Pos
) . t
, i
- 1 , std
: : make_pair ( ( * Pos
) . a
, ( * Pos
) . b
) ) ; Q
. erase ( Pos
) ; } } while ( Q
. size ( ) ) { Pos
= Q
. begin ( ) ; insert ( 1 , ( * Pos
) . t
, q
, std
: : make_pair ( ( * Pos
) . a
, ( * Pos
) . b
) ) ; Q
. erase ( Pos
) ; } dfs ( 1 ) ; return 0 ;
}
CF 1140 G
題目大意 ：nn n qq q n≤3?105,q≤6?105n\le 3·10^5,q\le6·10^5 n ≤ 3 ? 1 0 5 , q ≤ 6 ? 1 0 5 題解： fi,j,k,lf_{i,j,k,l} f i , j , k , l ? i,j∈[1,n],k,l∈[0,1]i,j\in[1,n],k,l\in[0,1] i , j ∈ [ 1 , n ] , k , l ∈ [ 0 , 1 ] ii i kk k 2j2^j 2 j ll l gig_i g i ? ii i gg g ff f O(nlogn)\mathcal O(nlogn) O ( n l o g n )
#include <bits/stdc++.h>
typedef long long ll
;
#define rg register
template
< typename T
> inline void read ( T
& x
) { char cu
= getchar ( ) ; x
= 0 ; bool fla
= 0 ; while ( ! isdigit ( cu
) ) { if ( cu
== '-' ) fla
= 1 ; cu
= getchar ( ) ; } while ( isdigit ( cu
) ) x
= x
* 10 + cu
- '0' , cu
= getchar ( ) ; if ( fla
) x
= - x
; }
template
< typename T
> inline void printe ( const T x
) { if ( x
>= 10 ) printe ( x
/ 10 ) ; putchar ( x
% 10 + '0' ) ; }
template
< typename T
> inline void print ( const T x
) { if ( x
< 0 ) putchar ( '-' ) , printe ( - x
) ; else printe ( x
) ; }
template
< typename T
> inline T
min ( const T a
, const T b
) { return a
< b
? a
: b
; }
template
< typename T
> inline T
max ( const T a
, const T b
) { return a
> b
? a
: b
; }
template
< typename T
> inline void mind ( T
& a
, const T b
) { a
= a
< b
? a
: b
; }
const int maxn
= 300001 , maxm
= 600001 ;
int n
, q
, dep
[ maxn
] ;
int head
[ maxn
] , nxt
[ maxm
] , tow
[ maxm
] , tmp
; ll vau1
[ maxm
] , vau2
[ maxm
] ;
inline void addb ( const int u
, const int v
, const ll w1
, const ll w2
)
{ tmp
++ ; nxt
[ tmp
] = head
[ u
] ; head
[ u
] = tmp
; tow
[ tmp
] = v
; vau1
[ tmp
] = w1
; vau2
[ tmp
] = w2
;
}
ll g
[ maxn
] ;
void dfs1 ( const int u
, const int fa
)
{ for ( rg
int i
= head
[ u
] ; i
; i
= nxt
[ i
] ) { const int v
= tow
[ i
] ; if ( v
== fa
) continue ; dep
[ v
] = dep
[ u
] + 1 ; dfs1 ( v
, u
) ; g
[ u
] = min ( g
[ u
] , vau1
[ i
] + vau2
[ i
] + g
[ v
] ) ; }
}
ll f
[ maxn
] [ 21 ] [ 2 ] [ 2 ] ; int p
[ maxn
] [ 21 ] ;
void dfs2 ( const int u
, const int fa
)
{ for ( rg
int i
= head
[ u
] ; i
; i
= nxt
[ i
] ) { const int v
= tow
[ i
] ; if ( v
== fa
) continue ; g
[ v
] = min ( g
[ v
] , vau1
[ i
] + vau2
[ i
] + g
[ u
] ) ; f
[ v
] [ 0 ] [ 0 ] [ 0 ] = vau1
[ i
] ; f
[ v
] [ 0 ] [ 0 ] [ 1 ] = g
[ v
] + vau2
[ i
] ; mind ( f
[ v
] [ 0 ] [ 0 ] [ 0 ] , f
[ v
] [ 0 ] [ 0 ] [ 1 ] + g
[ u
] ) ; mind ( f
[ v
] [ 0 ] [ 0 ] [ 1 ] , f
[ v
] [ 0 ] [ 0 ] [ 0 ] + g
[ u
] ) ; f
[ v
] [ 0 ] [ 1 ] [ 0 ] = g
[ v
] + vau1
[ i
] ; f
[ v
] [ 0 ] [ 1 ] [ 1 ] = vau2
[ i
] ; mind ( f
[ v
] [ 0 ] [ 1 ] [ 0 ] , f
[ v
] [ 0 ] [ 1 ] [ 1 ] + g
[ u
] ) ; mind ( f
[ v
] [ 0 ] [ 1 ] [ 1 ] , f
[ v
] [ 0 ] [ 1 ] [ 0 ] + g
[ u
] ) ; p
[ v
] [ 0 ] = u
; dfs2 ( v
, u
) ; }
}
ll dpu
[ 2 ] , dpv
[ 2 ] , dp
[ 2 ] ;
void C ( )
{ std
: : swap ( dpu
[ 0 ] , dpv
[ 0 ] ) ; std
: : swap ( dpu
[ 1 ] , dpv
[ 1 ] ) ;
}
void CP ( ll
* D
)
{ dp
[ 0 ] = D
[ 0 ] ; dp
[ 1 ] = D
[ 1 ] ;
}
ll
lca ( int u
, int v
)
{ if ( u
& 1 ) u
= ( u
+ 1 ) >> 1 , dpu
[ 0 ] = 0 , dpu
[ 1 ] = g
[ u
] ; else u
= ( u
+ 1 ) >> 1 , dpu
[ 0 ] = g
[ u
] , dpu
[ 1 ] = 0 ; if ( v
& 1 ) v
= ( v
+ 1 ) >> 1 , dpv
[ 0 ] = 0 , dpv
[ 1 ] = g
[ v
] ; else v
= ( v
+ 1 ) >> 1 , dpv
[ 0 ] = g
[ v
] , dpv
[ 1 ] = 0 ; if ( dep
[ u
] < dep
[ v
] ) std
: : swap ( u
, v
) , C ( ) ; const int DEP
= dep
[ u
] - dep
[ v
] ; for ( rg
int i
= 0 ; i
<= 20 ; i
++ ) if ( DEP
& ( 1 << i
) ) { CP ( dpu
) ; dpu
[ 0 ] = min ( dp
[ 0 ] + f
[ u
] [ i
] [ 0 ] [ 0 ] , dp
[ 1 ] + f
[ u
] [ i
] [ 1 ] [ 0 ] ) ; dpu
[ 1 ] = min ( dp
[ 0 ] + f
[ u
] [ i
] [ 0 ] [ 1 ] , dp
[ 1 ] + f
[ u
] [ i
] [ 1 ] [ 1 ] ) ; u
= p
[ u
] [ i
] ; mind ( dpu
[ 0 ] , dpu
[ 1 ] + g
[ u
] ) ; mind ( dpu
[ 1 ] , dpu
[ 0 ] + g
[ u
] ) ; } if ( u
== v
) return min ( dpu
[ 0 ] + dpv
[ 0 ] , dpu
[ 1 ] + dpv
[ 1 ] ) ; for ( rg
int i
= 20 ; i
>= 0 ; i
-- ) if ( p
[ u
] [ i
] != p
[ v
] [ i
] ) { CP ( dpu
) ; dpu
[ 0 ] = min ( dp
[ 0 ] + f
[ u
] [ i
] [ 0 ] [ 0 ] , dp
[ 1 ] + f
[ u
] [ i
] [ 1 ] [ 0 ] ) ; dpu
[ 1 ] = min ( dp
[ 0 ] + f
[ u
] [ i
] [ 0 ] [ 1 ] , dp
[ 1 ] + f
[ u
] [ i
] [ 1 ] [ 1 ] ) ; u
= p
[ u
] [ i
] ; mind ( dpu
[ 0 ] , dpu
[ 1 ] + g
[ u
] ) ; mind ( dpu
[ 1 ] , dpu
[ 0 ] + g
[ u
] ) ; CP ( dpv
) ; dpv
[ 0 ] = min ( dp
[ 0 ] + f
[ v
] [ i
] [ 0 ] [ 0 ] , dpv
[ 1 ] + f
[ v
] [ i
] [ 1 ] [ 0 ] ) ; dpv
[ 1 ] = min ( dp
[ 0 ] + f
[ v
] [ i
] [ 0 ] [ 1 ] , dpv
[ 1 ] + f
[ v
] [ i
] [ 1 ] [ 1 ] ) ; v
= p
[ v
] [ i
] ; mind ( dpv
[ 0 ] , dpv
[ 1 ] + g
[ v
] ) ; mind ( dpv
[ 1 ] , dpv
[ 0 ] + g
[ v
] ) ; } CP ( dpu
) ; dpu
[ 0 ] = min ( dp
[ 0 ] + f
[ u
] [ 0 ] [ 0 ] [ 0 ] , dp
[ 1 ] + f
[ u
] [ 0 ] [ 1 ] [ 0 ] ) ; dpu
[ 1 ] = min ( dp
[ 0 ] + f
[ u
] [ 0 ] [ 0 ] [ 1 ] , dp
[ 1 ] + f
[ u
] [ 0 ] [ 1 ] [ 1 ] ) ; u
= p
[ u
] [ 0 ] ; mind ( dpu
[ 0 ] , dpu
[ 1 ] + g
[ u
] ) ; mind ( dpu
[ 1 ] , dpu
[ 0 ] + g
[ u
] ) ; CP ( dpv
) ; dpv
[ 0 ] = min ( dp
[ 0 ] + f
[ v
] [ 0 ] [ 0 ] [ 0 ] , dpv
[ 1 ] + f
[ v
] [ 0 ] [ 1 ] [ 0 ] ) ; dpv
[ 1 ] = min ( dp
[ 0 ] + f
[ v
] [ 0 ] [ 0 ] [ 1 ] , dpv
[ 1 ] + f
[ v
] [ 0 ] [ 1 ] [ 1 ] ) ; v
= p
[ v
] [ 0 ] ; mind ( dpv
[ 0 ] , dpv
[ 1 ] + g
[ v
] ) ; mind ( dpv
[ 1 ] , dpv
[ 0 ] + g
[ v
] ) ; return min ( dpu
[ 0 ] + dpv
[ 0 ] , dpu
[ 1 ] + dpv
[ 1 ] ) ;
}
int main ( )
{ read ( n
) ; for ( rg
int i
= 1 ; i
<= n
; i
++ ) read ( g
[ i
] ) ; for ( rg
int i
= 1 ; i
< n
; i
++ ) { int u
, v
; ll w1
, w2
; read ( u
) , read ( v
) , read ( w1
) , read ( w2
) ; addb ( u
, v
, w1
, w2
) ; addb ( v
, u
, w1
, w2
) ; } dfs1 ( 1 , 0 ) ; p
[ 1 ] [ 0 ] = 1 , dfs2 ( 1 , 0 ) ; for ( rg
int j
= 1 ; j
<= 20 ; j
++ ) for ( rg
int i
= 1 ; i
<= n
; i
++ ) { const int P
= p
[ i
] [ j
- 1 ] ; p
[ i
] [ j
] = p
[ P
] [ j
- 1 ] ; const int T
= p
[ i
] [ j
] ; f
[ i
] [ j
] [ 0 ] [ 0 ] = min ( f
[ i
] [ j
- 1 ] [ 0 ] [ 0 ] + f
[ P
] [ j
- 1 ] [ 0 ] [ 0 ] , f
[ i
] [ j
- 1 ] [ 0 ] [ 1 ] + f
[ P
] [ j
- 1 ] [ 1 ] [ 0 ] ) ; f
[ i
] [ j
] [ 0 ] [ 1 ] = min ( f
[ i
] [ j
- 1 ] [ 0 ] [ 0 ] + f
[ P
] [ j
- 1 ] [ 0 ] [ 1 ] , f
[ i
] [ j
- 1 ] [ 0 ] [ 1 ] + f
[ P
] [ j
- 1 ] [ 1 ] [ 1 ] ) ; mind ( f
[ i
] [ j
] [ 0 ] [ 0 ] , f
[ i
] [ j
] [ 0 ] [ 1 ] + g
[ T
] ) ; mind ( f
[ i
] [ j
] [ 0 ] [ 1 ] , f
[ i
] [ j
] [ 0 ] [ 0 ] + g
[ T
] ) ; f
[ i
] [ j
] [ 1 ] [ 0 ] = min ( f
[ i
] [ j
- 1 ] [ 1 ] [ 0 ] + f
[ P
] [ j
- 1 ] [ 0 ] [ 0 ] , f
[ i
] [ j
- 1 ] [ 1 ] [ 1 ] + f
[ P
] [ j
- 1 ] [ 1 ] [ 0 ] ) ; f
[ i
] [ j
] [ 1 ] [ 1 ] = min ( f
[ i
] [ j
- 1 ] [ 1 ] [ 0 ] + f
[ P
] [ j
- 1 ] [ 0 ] [ 1 ] , f
[ i
] [ j
- 1 ] [ 1 ] [ 1 ] + f
[ P
] [ j
- 1 ] [ 1 ] [ 1 ] ) ; mind ( f
[ i
] [ j
] [ 1 ] [ 0 ] , f
[ i
] [ j
] [ 1 ] [ 1 ] + g
[ T
] ) ; mind ( f
[ i
] [ j
] [ 1 ] [ 1 ] , f
[ i
] [ j
] [ 1 ] [ 0 ] + g
[ T
] ) ; } read ( q
) ; while ( q
-- ) { int u
, v
; read ( u
) , read ( v
) ; print ( lca ( u
, v
) ) , putchar ( '\n' ) ; } return 0 ;
}
總結
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