hdu 1394 Minimum Inversion Number 归并求逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12107    Accepted Submission(s): 7388

Problem Description

The 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.

 

Input

The 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.

 

Output

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

 

Sample Input

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16
昨天做了几道逆序数的题，全用的树状数组解决的。晚上睡觉的时候发现，树状数组求逆序数，还是有很多麻烦的问题的，1，当序列的的离散程度比较大的大时候，，直接用树状数组，，会MLE，需要离散化序列，虽然不是太难，可模块一多，难免会出现一些比较隐晦的bug，在比赛的时候最忌讳这个了。2，当序列中有负数的时候，，可能我们就得重新选定坐标原点了，这个原点的坐标得根据数据的范围来选，如果题目说明的不清楚的时候，，就很坑了。
所以我决定再复习一下归并求逆序数，毕竟这个没什么限制，虽然速度不太怎么满意，但好在实用性比较强。
这道题就是给定一个序列，a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
让你把按照这些顺序的序列的逆序数全求出来：
a2, a3, ..., an, a1 (where m = 1)a3, a4, ..., an, a1, a2 (where m = 2)...an, a1, a2, ..., an-1 (where m = n-1)然后把最小的输出来。
没什么好说的，直接模板，
下面是代码：
#include <stdio.h>#define MAX 10000#define INF 100000000int b[MAX];int merge(int a[] , int start , int mid , int end){int i = start , j = mid+1 , k = start , count = 0;while(i<=mid && j<=end){if(a[i]<=a[j]){b[k++] = a[i++];}else{b[k++] = a[j++];count += mid-i+1;}}while(i<=mid){b[k++] = a[i++] ;}while(j<=end){b[k++] = a[j++] ;}for(i = start ; i <= end ; ++i){a[i] = b[i] ;}return count ;}int mergeSort(int a[],int start , int end){int sum = 0 ;if(start == end){return 0;}int mid = (start+end)>>1 ;sum +=mergeSort(a,start,mid) ;sum +=mergeSort(a,mid+1,end) ;sum +=merge(a,start,mid,end) ;return sum ;}int min(int a , int b){return a>b?b:a ;}int main(){int n ;while(~scanf("%d",&n)){int a[MAX],temp[MAX];for(int i = 0 ; i < n ; ++i){scanf("%d",&a[i]);temp[i] = a[i] ;}int sum = INF ,ans;sum = mergeSort(a,0,n-1) ;ans = sum ;for(int i = 0 ; i < n-1 ; ++i){sum += -temp[i]+ n-temp[i]-1;ans = min(sum,ans) ;}printf("%d\n",ans) ; }return 0;}

AC状态：