hdu 1394 最小逆序数

最新推荐文章于 2020-02-29 22:05:49 发布

西瓜君wtml

最新推荐文章于 2020-02-29 22:05:49 发布

阅读量312

点赞数

分类专栏：逆序数

本文链接：https://blog.youkuaiyun.com/q451792269/article/details/77532684

版权

逆序数专栏收录该内容

2 篇文章

订阅专栏

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

题意：

给定一个序列，每次可以把第一个数移动到尾部再形成一个序列，求这些序列逆序数的最小值。

题解：

先用归并排序求出原始序列逆序数。我们通过找规律可以发现
1 3 6 9 0 8 5 7 4 2 逆序数为22
3 6 9 0 8 5 7 4 2 1 逆序数为22-1+10-1-1=29
6 9 0 8 5 7 4 2 1 3 逆序数为29-3+10-1-3=32
所以可得下一个序列的逆序数为上一个序列的逆序数加上-num+n-1-num

代码：

#include <bits/stdc++.h>

using namespace std;
const int maxn = 5e3+10;
typedef long long LL;
LL cnt;
void merge_sort(LL A[],int x,int y,LL T[])
{
    if(y-x>1)
    {
        int m = x+(y-x)/2;  //划分
        int p=x,q=m,i=x;
        merge_sort(A,x,m,T);
        merge_sort(A,m,y,T);
        while(p<m||q<y)
        {
            if(q>=y||(p<m&&A[p]<=A[q]))
                T[i++]=A[p++];
            else
                {T[i++]=A[q++];cnt+=m-p;}
        }
        for(i=x;i<y;i++) A[i]=T[i];
    }
}

int main()
{
    LL A[maxn];
    LL T[maxn];
    LL C[maxn];
    int n;
    while(cin>>n)
    {
        cnt=0;
        memset(A,0,sizeof(A));
        memset(T,0,sizeof(T));
        memset(C,0,sizeof(C));
        for(int i=0;i<n;i++)
            {cin>>A[i];C[i]=A[i];}
        merge_sort(A,0,n,T);
        LL tmp=1<<30;
        for(int i=0;i<n;i++)
        {
           cnt=cnt-(C[i]-0)+(n-C[i]-1);
           tmp=min(cnt,tmp);
        }
        cout<<tmp<<endl;
    }
    return 0;
}