Minimum Inversion Number（HDU 1394）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10642    Accepted Submission(s): 6556

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
        

本題求最小逆序數。我用了3種方法（當然第3種是學别人的 ==）暴力、歸并、樹狀數組。。當然也可用線段樹，不過我不熟==。。注意本題求出最開始的逆序數後，可通過ans += (n-1-a[i])-a[i] 從0~n-1周遊求最小值即可（n為長度、a[i] 指每次變幻後的第一個數）。因為，首先要注意數字是0~n-1 ！！是以把a[i]放到最後面時，相當于後面有a[i]個數比a[i]小、有n-1-a[i]個數比a[i]大，結合起來就相當于增加了(n-1-a[i])-a[i]個逆序數，于是一直這樣操作後就可求出最小逆序數。

1.暴力：

#include<stdio.h>
int main()
{
    int n, i, j, A[5010];

    while(~scanf("%d", &n))
    {
        for(i=0; i<n; i++)
            scanf("%d", A+i);
        int cnt = 0;
        for(i=0; i<n; i++)
        {
            int t = 0;
            for(j=i+1; j<n; j++)
                if(A[i] > A[j]) t++;
            cnt += t;
        }
        int minn = cnt;
        for(i=0; i<n; i++)
        {
            cnt += (n-1-A[i])-A[i];
            if(cnt < minn) minn = cnt;
        }
        printf("%d\n", minn);
    }
    return 0;
}
           

2.利用歸并排序：

#include<stdio.h>
int cnt;
void merge_sort(int* A, int x, int y, int* temp)
{
    if(y-x > 1)
    {
        int m = x+(y-x)/2; //劃分
        int p = x, q = m, i = x;
        merge_sort(A, x, m, temp); //遞歸求解
        merge_sort(A, m, y, temp); //遞歸求解
        while(p < m || q < y)
        {
            if(q >= y || (p < m && A[p] <= A[q])) temp[i++] = A[p++]; 
            else temp[i++] = A[q++], cnt += m-p; //當取右邊的數時，即此時左邊的數比A[q]都小
        }
        for(i=x; i<y; i++) A[i] = temp[i];
    }
}

int main()
{
    int n, i, j, A[5010], temp[5010], B[5010];

    while(~scanf("%d", &n))
    {
        for(i=0; i<n; i++) //由于A、temp數組在歸并後都會改變，是以需要B數組。
            scanf("%d", A+i), B[i] = A[i];
        cnt = 0;
        merge_sort(A, 0, n, temp);
        int minn = cnt;
        for(i=0; i<n; i++)
        {
            cnt += (n-1-B[i])-B[i];
            if(cnt < minn) minn = cnt;
        }
        printf("%d\n", minn);
    }
    return 0;
}

           

3.樹狀數組:

首先，由于要用到樹狀數組，故需從1開始到n。是以也有A[i]++ 。分析所給資料序列中的5，它本來所在下标為6，現在為7。那麼sum[i]代表前面比5小的數的個數，前面共有7-1個數，故前面比5大的個數為i-1-sum[i]，即3個；于是再将A[i]當下标加到樹狀數組裡去；

#include<stdio.h>
#include<string.h>
int n, c[5010];

int lowbit(int x) {return x&(-x);}
void add(int i,int w)
{
    while(i <= n)
    {
        c[i] += w;
        i += lowbit(i);
    }
}
int sum(int i)
{
    int s = 0;
    while(i > 0)
    {
        s += c[i];
        i -= lowbit(i);
    }
    return s;
}

int main()
{
    int i, A[5010];

    while(~scanf("%d", &n))
    {
        memset(c, 0, sizeof(c));
       // memset(A, 0, sizeof(A));
        int ans = 0;
        for(i=1; i<=n; i++)
        {
            scanf("%d", A+i);
            A[i]++;
            //ans+=sum(n)-sum(A[i]);
            ans += i-1-sum(A[i]);
            add(A[i],1);
        }
        int minn = ans;
        for(i=1; i<=n; i++)
        {
            ans += n-A[i]-(A[i]-1); //注意上面序列++給這裡帶來的影響
            if(ans < minn) minn = ans;
        }
        printf("%d\n", minn);
    }
    return 0;
}