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
10 1 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 100000000 int 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: