hdu1394

先建立一个空的树；
在读入数据的过程中不断更新树，并计算出最初的序列的逆序对有多少；

还应知道每次将新的序列的第一个y[i]移到最后，sum将减少y[i]个，当然同时还会增加n-1-y[i];
例如：第一个是3 那么这个序列在3这个位置上必有0,1,2 即减少y[i]个；这三个逆序对；但以前不是3的逆序对
这时却是3的逆序对，即加上n-1-y[i]; 最后求出最小的值即可；



#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 5005;

#define LL(x) (x<<1)
#define RR(x) (x<<1|1)
#define MID(a, b) ((a+b)>>1)

int y[maxn], n;

struct node
{
    int lft, rht, value;
    int mid()
    {
        return MID(lft, rht);
    }
};

struct Segtree
{
    node tree[maxn*4];
    void build(int lft, int rht, int inf)
    {
        tree[inf].lft = lft;
        tree[inf].rht = rht;
        tree[inf].value = 0;
        if(lft != rht)
        {
            int mid = tree[inf].mid();
            build(lft, mid, LL(inf));
            build(mid+1, rht, RR(inf));
        }
    }
    void update(int pos, int inf)
    {
        int lft = tree[inf].lft, rht = tree[inf].rht;
        if(lft == rht) tree[inf].value++;
        else
        {
            int mid = tree[inf].mid();
            if(pos<=mid) update(pos, LL(inf));
            else update(pos, RR(inf));
            tree[inf].value = tree[LL(inf)].value + tree[RR(inf)].value;
        }
    }
    int query(int st, int en, int inf)
    {
        int lft = tree[inf].lft, rht = tree[inf].rht;
        if(st<=lft && rht<=en) return tree[inf].value;
        else
        {
            int mid = tree[inf].mid();
            int sum1 = 0, sum2 = 0;
            if(st <= mid) sum1 = query(st, en, LL(inf));
            if(mid < en) sum2 = query(st, en, RR(inf));
            return sum1+sum2;
        }
    }
}seg;

int main()
{
    while(scanf("%d", &n) != EOF)
    {
        seg.build(0, n-1, 1);
        int sum = 0;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &y[i]);
            seg.update(y[i], 1);
            if(y[i] != n-1) sum += seg.query(y[i]+1, n-1, 1);
        }
        int mi = sum;
        for(int i = 1; i <= n; i++)
        {
            sum -= y[i];
            sum += n-1-y[i];
            mi = min(mi, sum);
        }
        printf("%d\n", mi);
    }
    return 0;
}