LuoguP3521 [POI2011]ROT-Tree Rotations

题目

链接

题解

考虑对于一个子树有三种可能的逆序对：

1.在左子树中

2.在右子树中

3.跨过左右子树

显然对于交换左右子树的操作，只有第3种会受影响，且只有当前这个节点有关

所以对每个子树开个权值线段树即可，可以用线段树合并实现

#include<bits/stdc++.h>

using namespace std;
 
#define ll long long
#define re register
 
inline ll read()
{
    ll f = 1,x = 0;
    char ch;
    do
    {
        ch = getchar();
        if(ch == '-') f = -1;
    }while(ch<'0'||ch>'9');
    do
    {
        x = (x<<3) + (x<<1) + ch - '0';
        ch = getchar();
    }while(ch>='0'&&ch<='9');
    return f*x;
}

const int MAXN = 4e5 + 10;

int n;
int root = 0;
ll ans1,ans2,ans;
int lson[MAXN*40],rson[MAXN*40];
int sum[MAXN*40];
int cnt;

inline void update(int &rt,int l,int r,int k)
{
    if(!rt) rt = ++cnt;
    if(l == r)
    {
        sum[rt] += 1;
        return;
    }
    int mid = (l+r)>>1;
    sum[rt]++;
    if(k <= mid)
    update(lson[rt],l,mid,k);
    else update(rson[rt],mid+1,r,k);
    return;
}

inline void merge(int &l,int r)
{
    if(!l||!r)
    {
        l = l + r;
        return;
    }
    ans1 += (long long)sum[lson[l]] * (long long)sum[rson[r]]; 
    ans2 += (long long)sum[rson[l]] * (long long)sum[lson[r]];
    sum[l] += sum[r];
    merge(lson[l],lson[r]);
    merge(rson[l],rson[r]);
    return;
}

inline void dfs(int &rt)
{
    int val = read();
    int lc = 0,rc = 0;
    if(!val)
    {
        dfs(lc);dfs(rc);
        rt = lc;ans1=ans2=0;
        merge(rt,rc);
        ans += min(ans1,ans2);
    }
    else update(rt,1,n,val);
}

int main()
{
    n = read();
    dfs(root);
    cout << ans << endl;
}