hdu-1394-Minimum Inversion Number（线段树 || 树状数组）

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

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题意：有0～n-1组成的序列,然后进行下列操作,每次将最前面一个元素放到最后面去会得到一个序列,每得到一个序列都可得出该序列的逆序数（如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数）。要求求出最小的逆序数。

线段树

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#define N 5005
using namespace std;
int sum[N<<2], num[N];
void update(int p, int l, int r, int root)
{
    if (l == r)
    {
        sum[root]++;
        return ;
    }
    int m = (l+r)>>1;
    if (p <= m) update(p, l, m, root<<1);
    else    update(p, m+1, r, root<<1|1);
    sum[root] = sum[root<<1]+sum[root<<1|1];
}  
int query(int L, int R, int l, int r, int root)
{
    if (L <= l && r <= R)   return sum[root];
    int ans = 0;
    int m = (l+r)>>1;
    if (L <= m) ans += query(L, R, l, m, root<<1);
    if (R > m)  ans += query(L, R, m+1, r, root<<1|1);
    return ans;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    int i, j, n, Sum;
    while(cin >> n)
    {
        memset(sum, 0, sizeof(sum));
        Sum = 0;
        for (i = 0; i < n; i++)
        {
            cin >> num[i];
            Sum += query(num[i], n-1, 0, n-1, 1);
            update(num[i], 0, n-1, 1);
        }
        int ret = Sum;
        for (i = 0; i < n; i++)
        {
            Sum += n-num[i]*2-1;
            ret = min(ret, Sum);
        }
        cout << ret << endl;
    }
    return 0;
}

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树状数组

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm> 
#define N 5005
using namespace std;
int n, arr[N], num[N];
void update(int x, int d)
{
    while(x <= N)
    {
        arr[x] += d;
        x += x&-x;
    }
}
int sum(int x)
{
    int ret = 0;
    while(x > 0)
    {
        ret += arr[x];
        x -= x&-x;
    }
    return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    while(cin >> n)
    {
        memset(arr, 0, sizeof(arr));
        int ans = 0;
        for(int i = 1; i <= n; i++)
        {
            cin >> num[i];
            ans += sum(n+1)-sum(num[i]+1);
            update(num[i]+1, 1);
        }
        int tmp = ans;
        for (int i = 1; i <= n; i++)
        {
            tmp += n-1-num[i]*2;
            ans = min(ans, tmp);
        }
        cout << ans << endl;
    }
    return 0;
}