HDOJ 1394 Minimum Inversion Number

线段树，维护的是区间内整数的个数，每次插入 x 时，查询在 x 的前面比小的数的个数，并计算出比 x 大的数的个数 cnt[x] ，最后将 cnt 累加，即为逆序数；

将队首的数放到队尾后逆序数改变：n-1-a[i]， a[i] 为原始序列中第 i 个数的值；

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2012/7/15

之前的代码中，cnt[i] 是记录比第 i 个数大的的在它前面出现的数的个数，第二次写发现不需要。

# include <cstdio>
# include <cstring>

# define N 5000 + 5

int n, D, inv, a[N], sum[4 * N];

void update(int i)
{
    for (; i ^ 1; i >>= 1)
        sum[i >> 1] = sum[i] + sum[i ^ 1];
}

int query(int s, int t)
{
    int ans = 0;

    s += D-1, t += D+1;
    for (; s ^ t ^ 1; s >>= 1, t >>= 1)
    {
        if (~s & 0x1) ans += sum[s+1];
        if ( t & 0x1) ans += sum[t-1];
    }

    return ans;
}

void init(void)
{
    int tmp;

    inv = 0;
    for (D = 1; D < n+2; D <<= 1) ;
    memset(sum, 0, sizeof(sum[0])*D*2+2);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &tmp), ++tmp;
        a[i] = tmp;
        inv += i - 1 - query(1, tmp);
        sum[D+tmp] = 1, update(D+tmp);
    }
}

void solve(void)
{
    int Min;

    Min = inv;
    for (int i = 1; i < n; ++i)
    {
        inv = inv - (a[i]-1) + (n-a[i]);
        if (inv < Min) Min = inv;
    }
    printf("%d\n", Min);
}

int main()
{
    while (~scanf("%d", &n))
    {
        init();
        solve();
    }

    return 0;
}

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

# define N 50010

int n, D;
int sum[4 * N], cnt[N], a[N];

void update(int i)
{
    for (; i ^ 1; i >>= 1)
    {
        sum[i >> 1] = sum[i] + sum[i ^ 1];
    }
}

int query(int s, int t)
{
    int ret;
    
    ret = 0;
    s += D-1, t += D+1;
    for ( ; s ^ t ^ 1; s >>= 1, t >>= 1)
    {
        if (~s & 0x1) ret += sum[s+1];
        if ( t & 0x1) ret += sum[t-1];
    }

    return ret;
}

void init(void)
{
    int i;
    
    for (D = 1; D < n+2; D <<= 1) ;
    
    memset(sum, 0, sizeof(sum[0])*2*D+5);
    memset(cnt, 0, sizeof(cnt[0])*n+5);
    memset(a, 0, sizeof(a[0])*n+5);
    
    for (i = 1; i <= n; ++i)
    {
        scanf("%d", &a[i]);
        cnt[i] = i - 1 - query(0, a[i]);
        ++sum[a[i]+D], update(a[i]+D);
    }
}

void solve(void)
{
    int rev, i, min;

    rev = 0;
    for (i = 1; i <= n; ++i)
    {
        rev += cnt[i];
    }
    min = rev;
    for (i = 1; i <= n; ++i)
    {
        rev += n-1-2*a[i];
        if (min > rev) min = rev;
    }
    printf("%d\n", min);
}

int main()
{    
    while (~scanf("%d", &n))
    {
        init();
        solve();
    }

    return 0;
}

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