HDU_1394 Minimum Inversion Number(线段树)

　　/*设原序列的逆序数为sum，每往后移一个数，可以找到规律sum += n-2*num[i]-1;
(n为元素个数，num[i]为移动的数)；

求原序列的逆序数时可以用线段树，每输入一个数，先querry [输入的数，n]中元素的
个数，再把num[i]插入到线段树中。。。*/

#include <iostream>
#include <cstdio>
#include <cstring>
#define L(x) x << 1
#define R(x) x << 1 | 1

using namespace std;

const int N = 5005;

struct node
{
    int l, r;
    int sum;
}node[N*4];

int num[N];

void creat(int t, int l, int r)
{
    node[t].l = l;
    node[t].r = r;
    node[t].sum = 0;
    if(l == r)
        return;
    int mid = (l + r) >> 1;
    creat(L(t), l, mid);
    creat(R(t), mid + 1, r);
}

void updata(int t, int x)
{
    node[t].sum++;
    if(node[t].l == node[t].r)
        return;
    int mid = (node[t].l + node[t].r) >> 1;
    if(x <= mid)
        updata(L(t), x);
    else
        updata(R(t), x);
}

int query(int t, int l, int r)
{
    if(node[t].l >= l && node[t].r <= r)
        return node[t].sum;
    int mid = (node[t].l + node[t].r) >> 1;
    if(l > mid)
        return query(R(t), l, r);
    else if(r <= mid)
        return query(L(t), l, r);
    else
        return query(L(t), l, mid) + query(R(t), mid+1, r);
}

int main()
{
    //freopen("data.in", "r", stdin);

    int n, sum, i, min;
    while(scanf("%d", &n) != EOF)
    {
        creat(1, 0, n-1);
        sum = 0;
        for(i = 1; i <= n; i++)
        {
            scanf("%d", &num[i]);
            sum += query(1, num[i], n-1);
            updata(1, num[i]);
        }

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