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

Source: Internet
Author: User
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  ;} `

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.