HDU 3473 minimum sum (divide the tree, calculate the median, sum less than the median, and sum greater than the median)

Minimum Sum

Time Limit: 16000/8000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 1533 accepted submission (s): 345

Problem descriptionyou are given n positive integers, denoted as x0, X1... xN-1. then give you some intervals [L, R]. for each interval, you need to find a number X to make as small as possible!

 

Inputthe first line is an integer T (t <= 10), indicating the number of test cases. for each test case, an integer N (1 <= n <= 100,000) comes first. then comes n positive integers x (1 <= x <= 1,000,000,000) in the next line. finally, comes an integer Q (1 <= q <= 100,000), indicting there are Q queries. each query consists of two integers L, R (0 <= L <= r <n), meaning the interval you shoshould deal.

 

Outputfor the k-th test case, first output "Case # K:" in a separate line. then output Q lines, each line is the minimum value. output a blank line after every test case.

 

Sample input2 5 3 6 2 2 2 1 4 0 2 7 7 2 0 1 1 1

 

Sample outputcase #1: 6 4 case #2: 0 0

 

Authorstandy

 

Source2010ACM-ICPC multi-university training Contest (4) -- host by UESTC

 

Recommendzhengfeng is the partition tree. It is easy to know that when X is the median, the formula in the question is the smallest. Add sum [I] [J] to record the sum of numbers in layer I 1-j. Then, calculate the sum of values greater than the median and the sum of values smaller than the median. Code .

 /*  HDU 3473 minimum sumac g ++ 546 Ms 29000 K  */  # Include <Stdio. h> # Include < String . H> # Include <Iostream> # Include <Algorithm>Using   Namespace  STD;  Const   Int Maxn = 200010  ;  Int Tree [ 20  ] [Maxn];  Int  Sorted [maxn];  Int Toleft [ 20  ] [Maxn];  Long  Long Sum [ 20  ] [Maxn];  Void Build ( Int L, Int R, Int  Dep ){  If (L = R) {sum [Dep] [l] = Tree [Dep] [l];  Return  ;}  Int Mid = (L + r)>1  ;  Int Same = mid-L + 1 ; //  Same indicates the number of numbers equal to the center value and to the left.      For ( Int I = L; I <= r; I ++ ){  If (Tree [Dep] [I] < Sorted [Mid]) Same -- ; Sum [Dep] [I] = Tree [Dep] [I];  If (I> L) sum [Dep] [I] + = sum [Dep] [I- 1  ];}  Int LPOS = L;  Int Rpos = Mid + 1  ;  For ( Int I = L; I <= r; I ++ ){  If (Tree [Dep] [I] <sorted [Mid]) //  Go to left  {Tree [Dep +1 ] [LPOS ++] = Tree [Dep] [I];}  Else   If (Tree [Dep] [I] = sorted [Mid] & same> 0 ) //  Go to left  {Tree [Dep + 1 ] [LPOS ++] = Tree [Dep] [I]; same -- ;}  Else  //  Go to the right Tree [Dep + 1 ] [Rpos ++] = Tree [Dep] [I]; toleft [Dep] [I] = Toleft [Dep] [l- 1 ] + LPOS-L; //  Number of places from 1 to I on the left  } Build (L, mid, Dep + 1 ); //  Recursive build Build (Mid + 1 , R, DEP + 1  );}  Long  Long  Ans;  Int Query ( Int L, Int R, Int L, Int R, Int Dep, Int  K ){  If (L = r) Return  Tree [Dep] [l];  Int Mid = (L + r)>1  ;  Int CNT = toleft [Dep] [r]-toleft [Dep] [l- 1  ];  Int Ss = toleft [Dep] [l- 1 ]-Toleft [Dep] [l- 1  ];  Int EE = L-l- SS;  Int S = toleft [Dep] [r]-toleft [Dep] [l- 1  ];  Int E = r-L + 1 - S;  If (CNT> = K ){  If (E> 0  ){  If (EE> 0 ) Ans + = sum [Dep + 1 ] [Mid + E + EE]-sum [Dep + 1 ] [Mid + EE];  Else Ans + = sum [Dep +1 ] [Mid + E];}  //  L + number of numbers to the left before the query Interval          Int Newl = L + toleft [Dep] [l- 1 ]-Toleft [Dep] [l- 1  ];  //  The left endpoint + query interval will be split into the number on the left          Int Newr = newl + CNT- 1  ;  Return Query (L, mid, newl, newr, DEP +1 , K ); //  Note:  }  Else  {  If (S> 0  ){  If (Ss> 0 ) Ans-= sum [Dep + 1 ] [L + SS + S- 1 ]-Sum [Dep + 1 ] [L + SS- 1 ];  Else Ans-= sum [Dep + 1 ] [L + S- 1  ];}  //  Number of left-side splits after the R + interval          Int Newr = R + toleft [Dep] [r]- Toleft [Dep] [r];  //  Number of right endpoint minus the number of right-side partitions          Int Newl = newr-(R-l- CNT );  Return Query (Mid +1 , R, newl, newr, DEP + 1 , K-CNT ); //  Note:  }}  Int  Main (){  //  Freopen ("in.txt", "r", stdin );  //  Freopen ("out.txt", "W", stdout );      Int  T;  Int  N, m; scanf (  " % D  " ,& T );  Int  L, R;  Int Icase = 0  ;  While (T -- ) {Icase ++ ; Scanf (  "  % D  " ,& N ); For ( Int I = 1 ; I <= N; I ++ ) {Scanf (  "  % D  " , & Tree [ 0  ] [I]); sorted [I] = Tree [ 0  ] [I];} Sort (sorted + 1 , Sorted + 1 + N); Build ( 1 , N, 0  ); Printf (  "  Case # % d: \ n  "  , Icase); scanf (  "  % D  " ,& M );  While (M -- ) {Scanf (  "  % D  " , & L ,& R); L ++ ; R ++ ; Ans = 0  ;  Int Tt = query ( 1 , N, l, R, 0 , (R-l )/ 2 + 1  );  If (R-l + 1 ) % 2 = 0  ) {Ans -= TT;} printf (  "  % I64d \ n  "  , ANS);} printf (  "  \ N  "  );}  Return   0  ;}