线段树求逆序对hdu1394



学习自https://blog.youkuaiyun.com/metalseed/article/details/8039326

该题的数据从0~n-1排列，则需要的空间分配较少，

若不然，则可以将数据做离散处理

如1 4 5 6 可离散为1 2 3 4

先求出初始逆序数，再递推即可

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 5010;
int a[maxn];
int tree[maxn<<2];//表示该区间已经插入的数的个数
void build(int l, int r, int rt);
int query(int L, int R, int l, int r, int rt);
void update(int p, int l, int r, int rt);
void pushup(int rt);
int main()
{
    int i;
    int n;
    while(~scanf("%d", &n))
    {
        build(0, n - 1, 1);//建空树
        int cnt = 0;       //所求的初始逆序对
        for(i = 0; i < n; ++i)
        {
            scanf("%d", &a[i]);
            cnt += query(a[i], n - 1, 0, n - 1, 1);
            update(a[i], 0, n - 1, 1);
        }
        int ans = cnt;
        for(i = 0; i < n; ++i)
        {
            cnt += n - 1 - a[i] - a[i];//将a[i]从队首移至队尾，则逆序数减少a[i]个并且增加n - 1 - a[i]个
            ans = min(ans, cnt);
        }
        printf("%d\n", ans);
    }
    return 0;
}
void pushup(int rt)
{
    tree[rt] = tree[rt<<1] + tree[rt<<1|1];
}
void update(int p, int l, int r, int rt)
{
    if(l == r)
    {
        ++tree[rt];
        return;
    }
    int mid = (l + r)>>1;
    if(p <= mid)
        update(p, l, mid, rt<<1);
    else
        update(p, mid + 1, r, rt<<1|1);
    pushup(rt);
}
int query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R)
        return tree[rt];
    int mid = (l + r)>>1;
    int ret = 0;
    if(L <= mid)
        ret += query(L, R, l, mid, rt<<1);
    if(R >= mid + 1)
        ret += query(L, R, mid + 1, r, rt<<1|1);
    return ret;
}
void build(int l, int r, int rt)
{
    tree[rt] = 0;
    if(l == r) return;
    int mid = (l + r)>>1;
    build(l, mid, rt<<1);
    build(mid + 1, r, rt<<1|1);
}