hdu 1394 Minimum Inversion Number （树状数组求逆序对）

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.

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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

1 3 6 9 0 8 5 7 4 2

Sample Output

题目大意：

给定一个长为n的序列。求1..n，2..n..1，3..n..12，4..n..1..3，...这n种排列最小的逆序数。

1、树状数组

写这道题更多地是给自己一个树状数组的模板。建立在sum数组上的lowbit，add，getsum三个操作。

2、求逆序数

在初始全为0的长为n的数组上的操作。结构体排序。若有重复的值，注意排序的比较方式，应当是序号较大的排在后面，避免被记入逆序数。

3、本题技巧

0..n-1这n个数的序列。若第一个数为a，将其放至最后一位，则逆序数减少a，增加n-1-a，从而看做增加n-1-2a。

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=5000;
const int inf=1000000000;

int a[maxn+10];

struct tnode
{
    int num;
    int seq;
    bool operator<(const tnode& y) const
    {
        return num<y.num;
    }
};
tnode node[maxn+10];

int sum[maxn+10];

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

inline void add(int x,int val,int n)//向1..n序列的x位置加上val
{
    for(int i=x;i<=n;i+=lowbit(i))
        sum[i]+=val;
}

inline int getsum(int x)//1..x的和
{
    int ret=0;
    for(int i=x;i;i-=lowbit(i))
        ret+=sum[i];
    return ret;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        for(int i=1;i<=n;i++)
            node[i]=(tnode){a[i],i};
        sort(node+1,node+n+1);
        int ans0=0;
        memset(sum,0,sizeof(sum));
        for(int i=n;i>=1;i--)
        {
            ans0+=getsum(node[i].seq);
            add(node[i].seq,1,n);
        }
        int ans=ans0;
        for(int i=1;i<n;i++)
        {
            ans0=ans0+n-1-2*a[i];//0..n-1的排列逆序数规律
            ans=min(ans,ans0);
        }
        printf("%d\n",ans);
    }
    return 0;
}

View Code

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.

InputThe 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.
OutputFor 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

hdu 1394 Minimum Inversion Number （树状数组求逆序对）

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