Hdoj 1394 Minimum inversion number in order to find the smallest inverse of the loop string (violence && line tree)

Minimum Inversion number Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 14879 Accepted Submission (s): 9082


Problem DescriptionThe Inversion number of a given number sequence A1, A2, ..., the number of pairs (AI, aj) that SA Tisfy 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'll Obtain another sequence. There is 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 is 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 the 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 to a single line.

Sample Input

101 3 6 9 0 8 5 7 4 2

Sample Output

16

AC Code:

Brute Force Solution: 425ms

#include <stdio.h> #include <string.h> #include <math.h> #include <iostream> #include < algorithm> #define INF 0x7fffffff#define MAXN 10010#define Max (A, b) A>b?a:b#define min (A, b) a>b?b:ausing Namespace Std;int num[maxn];int Main () {int i,j,n;while (scanf ("%d", &n)!=eof) {int m;int cnt=0;for (i=0;i<n;i++) scanf ("%d", &num[i]); for (i=0;i<n;i++) {for (j=i+1;j<n;j++) {if (num[j]<num[i]) cnt++;}} M=cnt;for (i=0;i<n;i++) {cnt=cnt-num[i]+n-1-num[i];if (m>cnt) m=cnt;} printf ("%d\n", M);} return 0;}

See a lot of people use line tree to write, I also want to write, tomorrow!

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Hdoj 1394 Minimum inversion number in order to find the smallest inverse of the loop string (violence && line tree)