原题传送门
翻转就意味着Splay
然后肯定是先按顺序加入splay
首先找一找这道题有没有什么特点
发现每一个
p
i
pi
pi,在它前面的,两种情况:
- 权值比自己小,说明肯定在自己之前翻转到自己前面
- 权值比自己大,但位置在自己前面,这一点用splay维护翻转没毛病
总的来说,求pi很简单,其实只要把pi在二叉树中对应的点找到,旋到根,左儿子的size+1就是答案,其他的就是模板了
细节还是要注意一下
Code:
#include <bits/stdc++.h>
#define maxn 100010
#define inf 1 << 29
using namespace std;
int rt, sz, n, q[maxn];
int size[maxn], recy[maxn], son[maxn][2], tag[maxn], f[maxn];
struct node{
int v, id;
}a[maxn];
inline int read(){
int s = 0, w = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') w = -1;
for (; isdigit(c); c = getchar()) s = (s << 1) + (s << 3) + (c ^ 48);
return s * w;
}
bool cmp(node x, node y){ return x.v == y.v ? x.id < y.id : x.v < y.v; }
void update(int x){
if (x){
size[x] = recy[x];
if (son[x][0]) size[x] += size[son[x][0]];
if (son[x][1]) size[x] += size[son[x][1]];
}
}
int get(int x){ return son[f[x]][1] == x; }
void connect(int x, int y, int z){
if (x) f[x] = y;
if (y) son[y][z] = x;
}
void rotate(int x){
int fa = f[x], ffa = f[fa], m = get(x), n = get(fa);
connect(son[x][m ^ 1], fa, m);
connect(fa, x, m ^ 1);
connect(x, ffa, n);
update(fa); update(x);
}
void pushdown(int x){
if (tag[x]){
tag[x] = 0;
tag[son[x][0]] ^= 1;
tag[son[x][1]] ^= 1;
swap(son[x][0], son[x][1]);
}
}
void splay(int x, int goal){
int len = 0;
for (int i = x; i; i = f[i]) q[++len] = i;
for (int i = len; i; --i) pushdown(q[i]);
while (f[x] != goal){
int fa = f[x];
if (f[fa] != goal) rotate(get(x) == get(fa) ? fa : x);
rotate(x);
}
if (!goal) rt = x;
}
void build(int l, int r, int fa){
if (l > r) return;
int mid = (l + r) >> 1;
if (mid < fa) son[fa][0] = mid; else son[fa][1] = mid;
size[mid] = recy[mid] = 1, f[mid] = fa;
if (l == r) return;
build(l, mid - 1, mid);
build(mid + 1, r, mid);
update(mid);
}
int kth(int x){
int now = rt;
while (1){
pushdown(now);
if (size[son[now][0]] >= x) now = son[now][0]; else
if (size[son[now][0]] + 1 == x) return now; else
x -= size[son[now][0]] + 1, now = son[now][1];
}
}
void work(int l, int r){
l = kth(l), r = kth(r + 2);
splay(l, 0); splay(r, l);
tag[son[r][0]] ^= 1;
}
int main(){
n = read();
for (int i = 2; i <= n + 1; ++i){
a[i].v = read(), a[i].id = i;
}
a[1].id = 1, a[n + 2].id = n + 2;
a[1].v = -inf, a[n + 2].v = inf;
sort(a + 1, a + 3 + n, cmp);
build(1, n + 2, 0);
rt = (n + 3) >> 1;
for (int i = 2; i <= n; ++i){
splay(a[i].id, 0);
printf("%d ", size[son[rt][0]] + 1 - 1);
work(i - 1, size[son[rt][0]] + 1 - 1);
}
printf("%d\n", n);
return 0;
}