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  • hdu 1394 Minimum Inversion Number 逆序数/树状数组

    Minimum Inversion Number

    Time Limit: 1 Sec  Memory Limit: 256 MB

    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=1394

    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

    HINT

    题意

    这个区间可以变,就是可以把第一个数扔到最后去,然后这样变呀变,问这种变化下,最小的逆序数数是多少

    题解:

    啊,最大的数为n,把第一个数扔到最后,那么逆序数减少了num[i]-1,但是却增加了n-num[i],那就随便搞搞就好啦~  

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 100001
    #define mod 10007
    #define eps 1e-9
    const int inf=0x7fffffff;   //无限大
    /*
    inline ll read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    */
    //**************************************************************************************
    int d[maxn];
    int c[maxn];
    int n;
    int t;
    inline int read()
    {
        int x=0,f=1;char ch=getchar();
        while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    int lowbit(int x)
    {
        return x&-x;
    }
    
    void update(int x,int y)
    {
        while(x<=t)
        {
            d[x]+=y;
            x+=lowbit(x);
        }
    }
    int sum(int x)
    {
        int s=0;
        while(x>0)
        {
            s+=d[x];
            x-=lowbit(x);
        }
        return s;
    }
    int num[maxn];
    int main()
    {
        while(scanf("%d",&n)!=EOF)
        {
            memset(d,0,sizeof(d));
            memset(num,0,sizeof(num));
            memset(c,0,sizeof(c));
            //n=read();
            int ans=0;
            t=n;
            for(int i=0;i<n;i++)
            {
                num[i]=read();
                num[i]++;
                ans+=num[i]-sum(num[i]-1)-1;
                update(num[i],1);
            }
            int tmp=ans;
            for(int i=0;i<n;i++)
            {
                tmp+=n-1-2*num[i]+2;
                ans=min(tmp,ans);
            }
            cout<<ans<<endl;
        }
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4396831.html
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