HDU

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21660    Accepted Submission(s): 12952

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

 

题意：

        求一个数列的逆序数。不断把当前数列的第一个元素移到最后，构成新数列。求所有数列的最小逆序数。

题解：

       可以发现中间的元素不会改变， 由于n个数是0~n-1，那么将第一个数a[0]移到最后时，可以很容易的求出逆序数的改变量。假如a[0]=3，那么后面就有3个比它小的数，他们都要-1，即最终-3；移到末尾后，其前面有6个比它大的数，此时这一位的逆序数为6。那么就可以推出转移方程了：

dp[i]=dp[i-1]-a[i-1]+n-a[i-1]-1;

代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define inf 1<<29int n,a[5005],dp[5005];int cal(){    int cnt=0;    for(int i=1;i<n;i++)    {        for(int j=0;j<i;j++)            if(a[j]>a[i])                cnt++;    }    return cnt;}int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        memset(dp,0,sizeof dp);        dp[0]=cal();        int minn=dp[0];        for(int i=1;i<n;i++)        {            dp[i]=dp[i-1]-a[i-1]+n-a[i-1]-1;            minn=min(minn,dp[i]);        }        printf("%d\n",minn);    }    return 0;}