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CodeChef - DGCD——树链剖分+差分
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【題目描述】
You're given a tree on N vertices. Each vertex has a positive integer written on it, number on the ith vertex being vi. Your program must process two types of queries :1. Find query represented by F u v : Find out gcd of all numbers on the unique path between vertices u and v in the tree (both inclusive).2. Change query represented by C u v d : Add d to the number written on all vertices along the unique path between vertices u and v in the tree (both inclusive).
Input
First line of input contains an integer N denoting the size of the vertex set of the tree. Then follow N - 1 lines, ith of which contains two integers ai and bi denoting an edge between vertices ai and bi in the tree. After this follow N space separated integers in a single line denoting initial values vi at each of these nodes. Then follows a single integer Q on a line by itself, denoting the number of queries to follow. Then follow Q queries, each one on a line by itself. Each query is either a find query or a change query with format as given in problem statement. Note that all vertices are 0-based.
Output
For every find query, print the answer to that query in one line by itself.
Example
Input:
6
0 4
0 5
1 5
5 2
3 5
3 1 3 2 4 2
5
F 3 5
C 1 3 1
C 3 4 4
F 3 0
F 2 5
Output:
2
7
1
Constraints
1 <= N <= 50000
1 <= Q <= 50000
0 <= u, v <= N-1
1 <= vi <= 10^4
0 <= d <= 10^4
【題目分析】 gcd(x1,x2,x3...)=gcd(x1,x2?x1,x3?x2,...)gcd(x1,x2,x3...)=gcd(x1,x2-x1,x3-x2,...) g c d ( x 1 , x 2 , x 3 . . . ) = g c d ( x 1 , x 2 ? x 1 , x 3 ? x 2 , . . . ) gcd(x1,x2,x3...)=gcd(val1[x1],val2[x2..xn])gcd(x1,x2,x3...)=gcd(val1[x1],val2[x2..xn]) g c d ( x 1 , x 2 , x 3 . . . ) = g c d ( v a l 1 [ x 1 ] , v a l 2 [ x 2 . . x n ] ) 區間修改變成單點修改 ：對于每次區間修改，我們發現對于差分區間內部來講其實只改變了第一個的值，其他的都沒有改變（大家都增加了，減掉后值就沒有改變）。從重鏈往下，將兒子節點的值進行差分（父親節點-兒子節點）(理解這個十分重要) 。因此如果我們修改一個區間，重鏈的頭節點的值會增加val（因為沒有什么減他），修改的區間在這條鏈的最后的節點的兒子節點（如果有的話）會增加val（因為最后的節點的值增加了，兒子節點的值為最后節點減去兒子節點），如果區間沒有到達重鏈的頭節點，那么區間最前面的點的值會減小val（因為他的值應該是父親節點減去他，他的值增加而父親節點的值沒有變化所以他的值要減小val）。可能一開始還是有些難以理解（我理解了半天），可以對照著代碼。代碼應該還是比較清晰的。（注意區間修改是val1區間，單點修改是val2區間，區間查詢是val2區間，單點查詢是val1區間）【AC代碼】
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <climits>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std
; typedef long long ll
;
const int INF
= 0x3f3f3f3f ;
const int MAXN
= 5e4 + 5 ;
int head
[ MAXN
] , fa
[ MAXN
] , pos
[ MAXN
] , A
[ MAXN
] ;
int a
[ MAXN
] , sz
[ MAXN
] , son
[ MAXN
] , deep
[ MAXN
] , top
[ MAXN
] ;
struct node
{ int v
, next
;
} Edge
[ MAXN
<< 1 ] ;
int tot
= 0 , cnt
= 0 ;
int n
, m
;
int val1
[ MAXN
<< 2 ] , add
[ MAXN
<< 2 ] , val2
[ MAXN
<< 2 ] ; void AddEdge ( int u
, int v
)
{ tot
++ ; Edge
[ tot
] . v
= v
; Edge
[ tot
] . next
= head
[ u
] ; head
[ u
] = tot
;
} void dfs1 ( int u
, int f
)
{ fa
[ u
] = f
; deep
[ u
] = deep
[ f
] + 1 ; son
[ u
] = 0 ; sz
[ u
] = 1 ; for ( int i
= head
[ u
] ; i
; i
= Edge
[ i
] . next
) { int v
= Edge
[ i
] . v
; if ( v
!= f
) { dfs1 ( v
, u
) ; sz
[ u
] + = sz
[ v
] ; if ( sz
[ son
[ u
] ] < sz
[ v
] ) son
[ u
] = v
; } }
} void dfs2 ( int u
, int f
, int k
)
{ top
[ u
] = k
; pos
[ u
] = ++ cnt
; A
[ cnt
] = a
[ u
] ; if ( ! son
[ u
] ) return ; if ( son
[ u
] ) dfs2 ( son
[ u
] , u
, k
) ; for ( int i
= head
[ u
] ; i
; i
= Edge
[ i
] . next
) { int v
= Edge
[ i
] . v
; if ( v
!= f
&& v
!= son
[ u
] ) dfs2 ( v
, u
, v
) ; }
} void test ( )
{ for ( int i
= 1 ; i
< 13 ; i
++ ) { printf ( "%d " , add
[ i
] ) ; } printf ( "%d\n" , add
[ 13 ] ) ;
} void Build ( int k
, int l
, int r
)
{ add
[ k
] = 0 ; if ( l
== r
) { val1
[ k
] = A
[ l
] ; return ; } int mid
= ( l
+ r
) >> 1 ; Build ( k
<< 1 , l
, mid
) ; Build ( k
<< 1 | 1 , mid
+ 1 , r
) ;
} int gcd ( int a
, int b
)
{ return b
? gcd ( b
, a
% b
) : abs ( a
) ;
} void change_point ( int k
, int l
, int r
, int x
, int val
)
{ if ( l
== r
) { val2
[ k
] + = val
; return ; } int mid
= ( l
+ r
) >> 1 ; if ( x
<= mid
) change_point ( k
<< 1 , l
, mid
, x
, val
) ; else change_point ( k
<< 1 | 1 , mid
+ 1 , r
, x
, val
) ; val2
[ k
] = gcd ( val2
[ k
<< 1 ] , val2
[ k
<< 1 | 1 ] ) ;
} void pushdown ( int k
)
{ if ( add
[ k
] ) { add
[ k
<< 1 ] + = add
[ k
] ; add
[ k
<< 1 | 1 ] + = add
[ k
] ; val1
[ k
<< 1 ] + = add
[ k
] ; val1
[ k
<< 1 | 1 ] + = add
[ k
] ; add
[ k
] = 0 ; }
} void change_interval ( int k
, int l
, int r
, int L
, int R
, int val
)
{ if ( l
>= L
&& r
<= R
) { add
[ k
] + = val
; val1
[ k
] + = val
; return ; } int mid
= ( l
+ r
) >> 1 ; pushdown ( k
) ; if ( L
<= mid
) change_interval ( k
<< 1 , l
, mid
, L
, R
, val
) ; if ( R
> mid
) change_interval ( k
<< 1 | 1 , mid
+ 1 , r
, L
, R
, val
) ;
} void change_tree ( int u
, int v
, int w
)
{ while ( top
[ u
] != top
[ v
] ) { if ( deep
[ top
[ u
] ] < deep
[ top
[ v
] ] ) swap ( u
, v
) ; change_interval ( 1 , 1 , n
, pos
[ top
[ u
] ] , pos
[ u
] , w
) ; if ( son
[ u
] ) change_point ( 1 , 1 , n
, pos
[ son
[ u
] ] , w
) ; u
= fa
[ top
[ u
] ] ; } if ( deep
[ u
] < deep
[ v
] ) swap ( u
, v
) ; change_interval ( 1 , 1 , n
, pos
[ v
] , pos
[ u
] , w
) ; if ( son
[ u
] ) change_point ( 1 , 1 , n
, pos
[ son
[ u
] ] , w
) ; if ( son
[ fa
[ v
] ] == v
) change_point ( 1 , 1 , n
, pos
[ v
] , - w
) ;
} int query_interval ( int k
, int l
, int r
, int L
, int R
)
{ if ( l
>= L
&& r
<= R
) { return val2
[ k
] ; } int mid
= ( l
+ r
) >> 1 ; if ( R
<= mid
) return query_interval ( k
<< 1 , l
, mid
, L
, R
) ; else if ( L
> mid
) return query_interval ( k
<< 1 | 1 , mid
+ 1 , r
, L
, R
) ; else return gcd ( query_interval ( k
<< 1 , l
, mid
, L
, mid
) , query_interval ( k
<< 1 | 1 , mid
+ 1 , r
, mid
+ 1 , R
) ) ;
} int query_point ( int k
, int l
, int r
, int x
)
{ if ( l
== r
&& l
== x
) { return val1
[ k
] ; } int mid
= ( l
+ r
) >> 1 ; pushdown ( k
) ; if ( x
<= mid
) return query_point ( k
<< 1 , l
, mid
, x
) ; else return query_point ( k
<< 1 | 1 , mid
+ 1 , r
, x
) ;
} int query_tree ( int u
, int v
)
{ int ret
= 0 ; while ( top
[ u
] != top
[ v
] ) { if ( deep
[ top
[ u
] ] < deep
[ top
[ v
] ] ) swap ( u
, v
) ; if ( top
[ u
] != u
) ret
= gcd ( ret
, query_interval ( 1 , 1 , n
, pos
[ son
[ top
[ u
] ] ] , pos
[ u
] ) ) ; ret
= gcd ( ret
, query_point ( 1 , 1 , n
, pos
[ top
[ u
] ] ) ) ; u
= fa
[ top
[ u
] ] ; } if ( deep
[ u
] < deep
[ v
] ) swap ( u
, v
) ; if ( u
!= v
) ret
= gcd ( ret
, query_interval ( 1 , 1 , n
, pos
[ son
[ v
] ] , pos
[ u
] ) ) ; ret
= gcd ( ret
, query_point ( 1 , 1 , n
, pos
[ v
] ) ) ; return ret
;
} void Read ( )
{ int u
, v
; scanf ( "%d" , & n
) ; for ( int i
= 1 ; i
< n
; i
++ ) { scanf ( "%d%d" , & u
, & v
) ; AddEdge ( u
+ 1 , v
+ 1 ) ; AddEdge ( v
+ 1 , u
+ 1 ) ; } for ( int i
= 1 ; i
<= n
; i
++ ) scanf ( "%d" , & a
[ i
] ) ;
} void Init ( )
{ dfs1 ( 1 , 0 ) ; dfs2 ( 1 , 0 , 1 ) ; Build ( 1 , 1 , n
) ; for ( int i
= 1 ; i
<= n
; i
++ ) { if ( son
[ i
] ) change_point ( 1 , 1 , n
, pos
[ son
[ i
] ] , a
[ i
] - a
[ son
[ i
] ] ) ; }
} void Solve ( )
{ char cmd
[ 10 ] ; int u
, v
, w
; scanf ( "%d" , & m
) ; while ( m
-- ) { scanf ( "%s" , cmd
) ; if ( cmd
[ 0 ] == 'F' ) { scanf ( "%d%d" , & u
, & v
) ; printf ( "%d\n" , query_tree ( u
+ 1 , v
+ 1 ) ) ; } else { scanf ( "%d%d%d" , & u
, & v
, & w
) ; change_tree ( u
+ 1 , v
+ 1 , w
) ; } }
} int main ( )
{ Read ( ) ; Init ( ) ; Solve ( ) ; return 0 ;
}
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