Description

Solution

對於每個節點有一棵權值線段樹，向上遞迴時合併同時計算逆序對即可。

Source

/************************************************
 * Au: Hany01
 * Date: Mar 22nd, 2018
 * Prob: [BZOJ2122][POI2011] Tree Rotations
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>
 


using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
 

#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir firct
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
 

#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 4000005;

int n, lc[maxn * 2], rc[maxn * 2], ret, tr[maxn], cnt;
LL cnt1, cnt2, Ans;

#define mid ((l + r) >> 1)
int build(int l, int r, int x)
{
    int now = ++ cnt;
    if (l < r) {
        if (x <= mid) lc[now] = build(l, mid, x);
        else rc[now] = build(mid + 1, r, x);
    }
    tr[now] = 1;
    return now;
}

int merge(int x, int y)
{
    if (!x || !y) return x + y;
    cnt1 += (LL)tr[lc[x]] * (LL)tr[rc[y]], cnt2 += (LL)tr[rc[x]] * (LL)tr[lc[y]];
    lc[x] = merge(lc[x], lc[y]), rc[x] = merge(rc[x], rc[y]);
    tr[x] = tr[lc[x]] + tr[rc[x]];
    return x;
}

int dfs()
{
    register int tmp = read(), x, y, ret;
    if (tmp) return build(1, n, tmp);
    x = dfs(), y = dfs();
    cnt1 = cnt2 = 0;
    ret = merge(x, y);
    Ans += min(cnt1, cnt2);
    return ret;
}

int main()
{
#ifdef hany01
    File("bzoj2212");
#endif

    n = read();

    dfs();

    printf("%lld\n", Ans);

    return 0;
}
//香臉半開嬌旖旎。當庭際。玉人浴出新妝洗。
//    -- 李清照《漁家傲·雪裡已知春信至》