HDU 1166 enemy deployment (tree array)

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1166

Analysis: The first time I came into contact with the line segment tree, I read some information online and found that this question can be solved by replacing the line segment tree with a tree array. It seems that it is more convenient to use, as you can see from the tree array, here is a tree array of data that I extracted from the Internet. The tree array is not very complete, but it is enough for beginners. There are basic templatesCode.

This is a Chinese question. The question is very simple. It is a typical tree array and line segment tree question.

Tree array Solution

# Include <iostream> # Include <Cstdio> # Include <Cstring> # Include <Algorithm> # Define Maxn 50005 Using   Namespace  STD;  Int A [maxn], C [maxn];  Int  N;  //  Tree array template  Int Low_bit ( Int  X ){  Return X &(- X );}  Void Modify ( Int N, Int  Value ){  While (N <=N) {C [N] + = Value; n + = Low_bit (n );}}  Int Sum ( Int  N ){  Int Sum = 0  ;  While (N> 0  ) {Sum + = C [N]; n -= Low_bit (n );} Return  SUM ;}  Int Query ( Int I, Int  J ){  Return Sum (j)-sum (I- 1  );}  Int  Main (){  Int  T, x, y; scanf (  "  % D  " ,&T );  Char Order [ 10  ];  For ( Int I = 1 ; I <= T; I ++ ) {Printf (  "  Case % d: \ n  "  , I); scanf (  "  % D  " ,&N); memset (,  0 , Sizeof  (A); memset (C,  0 , Sizeof  (C ));  For ( Int J = 1 ; J <= N; j ++ ) {Scanf (  "  % D  " ,&A [J]); Modify (J, a [J]);}  While (Scanf ( "  % S  " , Order) & strcmp (order, "  End  " )! = 0  ) {Scanf (  "  % D  " , & X ,& Y );  If (Strcmp (order, "  Query  " ) = 0  ) Printf (  "  % D \ n  "  , Query (x, y ));  Else   If (Strcmp (order, "  Add  " ) = 0 ) Modify (x, y );  Else   If (Strcmp (order, "  Sub  " ) = 0  ) Modify (X, - Y );}}  Return   0  ;}

Line Segment tree solution:
 

# Include <iostream> # Include <Cstdio> # Include <Cstring> # Include <Algorithm> # Define Maxn 50005 Using   Namespace  STD; typedef  Struct  Node {  Int LD, RD; //  Left and right boundary      Struct Node * LC, * RC; //  Left and right subtree      Int Value;} node, tree;  Int  Num [maxn];  Int  N; tree * Create ( Int A, Int  B ){  Int  Mid; Node * Node = (node *) malloc ( Sizeof  (Node); Node -> LD = A; Node -> RD = B; If (B-A> = 1  ) {Mid = (A + B )/ 2  ; Node -> Lc = Create (A, mid); Node -> Rc = create (Mid + 1  , B); Node -> Value = node-> LC-> value + node-> RC-> Value ;}  Else Node-> value = num [node-> ld]; //  You can also write node-> value = A [node-> RD];     Return  Node ;}  Int Query (node * node, Int X, Int  Y ){  If (Node-> LD = x & node-> RD = y) Return Node-> Value;  Int Mid = (node-> LD + node-> rd )/ 2  ;  If (Mid> = y) Return Query (node-> LC, x, y );  Else   If (Mid <X) Return Query (node-> RC, x, y );  Else   Return Query (node-> LC, X, mid) + query (node-> RC, Mid + 1  , Y );}  Void Modify (node * node, Int X, Int  V ){  If (Node-> LD = node-> Rd) {Node -> Value + = V;  Return  ;}  If (Node-> LC-> RD> = x) Modify (node-> LC, X, V );  Else   If (Node-> RC-> LD <= x) Modify (node-> RC, X, V); Node -> Value = node-> LC-> value + node-> RC-> Value ;}  Int  Main (){ Int  T, x, y, value; Node * Root; scanf (  "  % D  " ,& T );  Char Order [ 10  ];  For ( Int I = 1 ; I <= T; I ++ ) {Printf (  " Case % d: \ n  "  , I); scanf (  "  % D  " ,& N );  For ( Int J = 1 ; J <= N; j ++ ) Scanf (  "  % D  " ,& Num [J]); root = Create (1  , N );  While (Scanf ( "  % S  " , Order) & strcmp (order, "  End  " )! = 0  ) {Scanf (  "  % D  " , & X ,& Y ); If (Strcmp (order, "  Query  " ) = 0  ) Printf (  "  % D \ n  "  , Query (root, x, y ));  Else   If (Strcmp (order, "  Add  " ) = 0 ) Modify (root, x, y );  Else   If (Strcmp (order, "  Sub  " ) = 0  ) Modify (root, X, - Y );}}  Return   0  ;}