hdu1394 Minimum Inversion Number(最小逆序数)
2024-09-28 19:02:45
Minimum Inversion Number
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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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.
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
Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
先百度逆序数:在一个排列中。假设一对数的前后位置与大小顺序相反,即前面的数大于后面的数,那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数。逆序数为偶数的排列称为偶排列;逆序数为奇数的排列称为奇排列。
如2431中,21。43,41,31是逆序,逆序数是4,为偶排列。
对于这个题求把第一个数放到最后一个数的最小逆序数,对于原序列而言,假设把第一个数放到最后一个数,逆序列添加n-num[0]+1,逆序列降低
num[0].
所以这道题:
#include <stdio.h>
int main()
{
int num[5005],n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
scanf("%d",&num[i]);
int sum=0,temp;
for(int i=0;i<n;i++)
for(int j=i+1;j<n;j++)
if(num[i]>num[j]) sum++;
temp=sum;
for(int i=n-1;i>=0;i--)
{
temp-=n-1-num[i];
temp+=num[i];
if(temp<sum)
sum=temp;
}
printf("%d\n",sum);
}
return 0;
}
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