题意：有一个n个整数的排列，这n个整数就是0,1,2,3...n-1这n个数(但不一定按这个顺序给出)。现在先计算一下初始排列的逆序数，然后把第一个元素a1放到an后面去，形成新排列a2 a3 a4...an a1，然后再求这个排列的逆序数。继续执行类似操作(一共要执行n-1次)直到产生排列an a1 a2...an-1为止。计算上述所有排列的逆序数，输出最小逆序数。
思路：首先由于给出的序列是0到n-1，所以要+1，因为树状数组第0项是无法使用的。然后先求出一开始序列的逆序数，现在问题是怎么求题目要求的经过n-1次变换之后的最小逆序数呢？
我们假设一开始的数列是 5 1 2 3 4 6 7 8 9 10，将5放到最后，那么就有4个比5小的逆序数减少1，有10-5个比5大的逆序数增加1，那么变换之后的序列数的公式就推出来了，就是ans-(ai-1)+(n-ai)，自己推一下看看，那么这道题就解决了。
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <set>
#include <ctime>
#include <cmath>
#include <cctype>
using namespace std;
#define maxn 5000+100
#define LL long long
int cas=1,T;
int c[maxn];
int lowbit(int x)
{
return x&(-x);
}
int sum (int x)
{
int ans = 0;
while (x)
{
ans+=c[x];
x-=lowbit(x);
}
return ans;
}
void add(int x,int v)
{
while (x<=maxn)
{
c[x]+=v;
x+=lowbit(x);
}
}
int a[maxn];
int main()
{
int n;
while (scanf("%d",&n)!=EOF )
{
int anss = 0;
memset(c,0,sizeof(c));
for (int i = 1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]++;
anss+=sum(n)-sum(a[i]);
add(a[i],1);
}
int ans = anss;
for (int i = 1;i<=n;i++)
{
anss +=n-2*a[i]+1;
ans = min(ans,anss);
}
printf("%d\n",ans);
}
//freopen("in","r",stdin);
//scanf("%d",&T);
//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
return 0;
}
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