HDU 1394 minimum inversion number (number of reverse orders, line segment tree or tree array)

Minimum inversion number

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 4647 accepted submission (s): 2809

Problem descriptionthe inversion number of a given number sequence A1, A2,..., an is the number of pairs (AI, AJ) that satisfy I <j and AI> AJ.

For a given sequence of numbers A1, A2 ,..., an, if we move the first m> = 0 numbers to the end of the seqence, we will obtain another sequence. there are totally N such sequences as the following:

A1, A2,..., An-1, an (where m = 0-the initial seqence)
A2, A3,..., An, A1 (where M = 1)
A3, A4,..., An, A1, A2 (where m = 2)
...
An, A1, A2,..., An-1 (where M = N-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

Inputthe input consists of a number of test cases. each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the N integers from 0 to n-1.

 

Outputfor each case, output the minimum inversion number on a single line.

 

Sample input10 1 3 6 9 0 8 5 7 4 2

 

Sample output16

 

Authorchen, Gaoli

 

Sourcezoj monthly, January 2003

 

Recommendignatius. l this question is to find the minimum value of the number of reverse orders after the cyclic shift. In fact, it is mainly to find the number of reverse sequences. There are many methods to calculate the number of reverse orders, which can be obtained by merging and sorting. You can also use a tree array and line segment tree to calculate the number of reverse orders. After the number of reverse orders is obtained, you can directly move the first number to the last number of reverse orders. For example, if the original number of reverse orders is ans, move a [0] to the end to reduce the number of reverse orders. A [0], at the same time, increase the number of reverse orders N-A [0]-1 is ANS-A [0] + N-A [0]-1; as long as I gets the minimum value from a 0-n-1 loop. Line Segment tree practices:

 /*  HDU 1394g ++ 78 Ms 280 K  */  # Include <Stdio. h> # Include < String . H> # Include <Algorithm> Using   Namespace  STD; Const   Int Maxn = 5050  ;  Struct  Node {  Int  L, R;  Int  SUM;} segtree [maxn * 3  ];  Void Build ( Int I, Int L, Int R) {segtree [I]. L = L; segtree [I]. r = R;  If (L = R) {segtree [I]. Sum = 0  ;  Return  ;}  Int Mid = (L + r)> 1  ; Build (I < 1  , L, mid); Build (I <1 ) | 1 , Mid + 1  , R); segtree [I]. Sum = 0  ;}  Void Add ( Int I, Int T, Int  Val) {segtree [I]. Sum + = Val;  If (Segtree [I]. L = Segtree [I]. R ){ Return  ;}  Int Mid = (segtree [I]. L + segtree [I]. R)> 1  ;  If (T <= mid) add (I < 1  , T, Val );  Else Add (I < 1 ) | 1  , T, Val );}  Int Sum ( Int I,Int L, Int  R ){  If (Segtree [I]. L = L & segtree [I]. r = R)  Return  Segtree [I]. sum;  Int Mid = (segtree [I]. L + segtree [I]. R)> 1  ;  If (R <= mid) Return Sum (I < 1  , L, R ); Else   If (L> mid) Return Sum (I < 1 ) | 1  , L, R );  Else   Return Sum (I < 1 , L, mid) + sum (I < 1 ) | 1 , Mid + 1  , R );}  Int  A [maxn]; Int  Main (){  //  Freopen ("in.txt", "r", stdin );  //  Freopen ("out.txt", "W", stdout );      Int  N;  While (Scanf ( "  % D  " , & N )! = EOF) {build (  1 , 0 , N-1  );  For ( Int I = 0 ; I <n; I ++ ) Scanf (  "  % D  " ,& A [I]);  Int Ans = 0  ;  For ( Int I =0 ; I <n; I ++ ) {Ans + = Sum ( 1 , A [I], n- 1  ); Add (  1 , A [I], 1  );}  Int Min = Ans;  For ( Int I = 0 ; I <n; I ++ ) {Ans -= A [I]; //  Descending Order Ans + = n-A [I]- 1  ;  If (ANS <min) min = Ans;} printf (  "  % D \ n  "  , Min );}  Return   0  ;} 

 

 

Tree array:

 /* HDU 1394ac g ++ 46 Ms 252 K  */  # Include <Stdio. h> # Include < String . H> # Include <Algorithm> Using   Namespace  STD;  Const   Int Maxn = 5050  ;  Int  C [maxn]; Int  N;  Int Lowbit ( Int  X ){  Return X &(- X );}  Void Add ( Int I, Int  Val ){  While (I <= N) {C [I] + = Val; I + =Lowbit (I );}}  Int Sum ( Int  I ){  Int S = 0  ;  While (I> 0  ) {S + = C [I]; I -= Lowbit (I );}  Return  S ;}  Int A [maxn];  Int  Main (){  While (Scanf ( "  % D  " , & N )! = EOF ){  Int Ans = 0  ; Memset (C,  0 , Sizeof  (C ));  For (Int I = 1 ; I <= N; I ++ ) {Scanf (  "  % D  " ,& A [I]); A [I] ++ ; Ans + = Sum (N )- Sum (A [I]); add (A [I],  1  );}  Int Min = Ans;  For (Int I = 1 ; I <= N; I ++ ) {Ans + = N-A [I]-(A [I]- 1  );  If (ANS <min) min = Ans;} printf (  "  % D \ n  "  , Min );}  Return   0  ;}