HDU1394-Minimum Inversion Number

最新推荐文章于 2021-08-20 18:20:00 发布

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最新推荐文章于 2021-08-20 18:20:00 发布

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分类专栏： HDU ----暴力 ----树状数组

本文链接：https://blog.youkuaiyun.com/a664607530/article/details/52549066

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----暴力

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----树状数组

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本文介绍了一种算法问题——寻找给定序列通过不同轮转方式所能得到的序列中逆序对数量的最小值。提供了两种解决方案，一种使用树状数组进行优化计算，另一种采用直接比较的方式实现。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18396 Accepted Submission(s): 11168

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

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

题意：给定一个长度为n的数组，那么每次将第一个元素放在最后一个能形成n个长度为n的数组，求出所有情况中逆序对最少的个数。

解题思路：设当前第一个元素为x，那么比它小的有x个元素，比它大的有n-x-1个元素，将它放到数组最后对逆序对的变化是n-2*x-1个。那么我们只要枚举这个分开的位置m就行了。

树状数组：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAXN=5050;
int f[MAXN],a[MAXN];;
int n;

int lowbit(int x)
{
    return x&(-x);
}

void add(int x,int val)
{
    while(x<=n)
    {
        f[x]+=val;
        x+=lowbit(x);
    }
}

int getsum(int x)
{
    int s=0;
    while(x>0)
    {
        s+=f[x];
        x-=lowbit(x);
    }
    return s;
}

int main()
{
    while(~scanf("%d",&n))
    {
        int ans=0;
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            ans+=getsum(n)-getsum(a[i]);
            add(a[i],1);
        }
        int mi=ans;
        for(int i=1;i<=n;i++)
        {
            ans+=n-a[i]-(a[i]-1);
            if(ans<mi) mi=ans;
        }
        printf("%d\n",mi);
    }
    return 0;
}

暴力：

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

int main()
{
    int a[5009],sum[5009];
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        memset(sum,0,sizeof sum);
        for(int i=0;i<n-1;i++)
        {
            for(int j=i+1;j<n;j++)
                if(a[j]<a[i]) sum[i]++;
        }
        int ans=0;
        for(int i=0;i<n;i++)
            ans+=sum[i];
        int t=ans;
        for(int i=0;i<n-1;i++)
        {
            t=t-sum[i]+n-1-sum[i];
            for(int j=i+1;j<n;j++)
                if(a[i]<a[j]) sum[j]++;
            if(t<ans) ans=t;
        }
        printf("%d\n",ans);
    }
    return 0;
}