【HNOI2017】单旋
题面
题解
trajan的spaly是O(1)的(逃
通过看题解手玩发现只要将最值的点放到树根,其他的父子关系不需要变。
于是想到动态连边和断边的数据结构:\(\mathrm{LCT}\),于是用\(\mathrm{LCT}\)维护\(\mathrm{splay}\)。
这样后面的四个操作就解决了。
第一个操作中,第一个点要么接在前驱的右儿子,要么接在后继的左儿子。
于是再开个set求一下前驱后继即可。
代码
很多可以复制粘贴,所以也不算很长。
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<set>
#include<map>
#define RG register
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define clear(x, y) memset(x, y, sizeof(x))
inline int read()
{
int data = 0, w = 1; char ch = getchar();
while(ch != '-' && (!isdigit(ch))) ch = getchar();
if(ch == '-') w = -1, ch = getchar();
while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
return data * w;
}
const int maxn(200010);
int m, root, fa[maxn], son[2][maxn], stk[maxn], top;
int size[maxn], rev[maxn], cur, val[maxn], lson[maxn], rson[maxn], Fa[maxn];
inline bool isroot(int x) { return son[0][fa[x]] != x && son[1][fa[x]] != x; }
inline int getSon(int x) { return son[1][fa[x]] == x; }
inline void update(int x) { size[x] = size[son[0][x]] + size[son[1][x]] + 1; }
inline void pushdown(int x)
{
if(!rev[x]) return;
rev[son[0][x]] ^= 1;
rev[son[1][x]] ^= 1;
rev[x] = 0, std::swap(son[0][x], son[1][x]);
}
inline void rotate(int x)
{
int f = fa[x], g = fa[f], l = getSon(x), r = l ^ 1;
if(!isroot(f)) son[getSon(f)][g] = x;
fa[x] = g, fa[f] = x, fa[son[r][x]] = f;
son[l][f] = son[r][x], son[r][x] = f;
update(f), update(x);
}
void splay(int x)
{
stk[top = 1] = x;
for(int i = x; !isroot(i); i = fa[i]) stk[++top] = fa[i];
while(top) pushdown(stk[top--]);
for(; !isroot(x); rotate(x)) if(!isroot(fa[x]))
rotate(getSon(fa[x]) ^ getSon(x) ? x : fa[x]);
}
void access(int x)
{
for(int t = 0; x; t = x, x = fa[x])
splay(x), son[1][x] = t, update(x);
}
inline void makeroot(int x) { access(x); splay(x); rev[x] ^= 1; }
inline void split(int x, int y) { makeroot(x); access(y); splay(y); }
inline int getdep(int x) { split(x, root); return size[root]; }
inline void link(int x, int y) { makeroot(x); fa[x] = y; }
inline void cut(int x, int y)
{
split(x, y); if(son[0][y] == x) son[0][y] = 0;
fa[x] = 0;
}
std::map<int, int> map;
std::set<int> set;
void insert(int x)
{
int now = map[x] = ++cur;
if(set.empty()) return (void) (set.insert(x), root = now, puts("1"));
std::set<int>::iterator it = set.upper_bound(x);
if(it == set.end() || lson[map[*it]])
{
--it; int node = map[*it]; link(now, node);
Fa[now] = node, rson[node] = now;
}
else
{
int node = map[*it]; link(now, node);
Fa[now] = node, lson[node] = now;
}
set.insert(x); printf("%d\n", getdep(now));
}
void splay_min()
{
int x = map[*set.begin()];
if(root == x) return (void) (puts("1"));
printf("%d\n", getdep(x)); cut(x, Fa[x]);
if(rson[x]) cut(x, rson[x]);
link(x, root); if(rson[x]) link(Fa[x], rson[x]);
lson[Fa[x]] = rson[x]; if(rson[x]) Fa[rson[x]] = Fa[x];
Fa[x] = 0, Fa[root] = x, rson[x] = root; root = x;
}
void splay_max()
{
int x = map[*set.rbegin()];
if(root == x) return (void) (puts("1"));
printf("%d\n", getdep(x)); cut(x, Fa[x]);
if(lson[x]) cut(x, lson[x]);
link(x, root); if(lson[x]) link(lson[x], Fa[x]);
rson[Fa[x]] = lson[x]; if(lson[x]) Fa[lson[x]] = Fa[x];
Fa[x] = 0, Fa[root] = x, lson[x] = root; root = x;
}
void del_min()
{
int x = map[*set.begin()];
if(root == x)
{
puts("1"); if(rson[x]) cut(x, rson[x]);
Fa[rson[x]] = 0; root = rson[x];
set.erase(set.begin()); return;
}
printf("%d\n", getdep(x)); cut(x, Fa[x]);
if(rson[x]) cut(x, rson[x]), link(Fa[x], rson[x]);
lson[Fa[x]] = rson[x]; if(rson[x]) Fa[rson[x]] = Fa[x];
set.erase(set.begin());
}
void del_max()
{
int x = map[*set.rbegin()];
if(root == x)
{
puts("1"); if(lson[x]) cut(x, lson[x]);
Fa[lson[x]] = 0; root = lson[x];
set.erase(--set.end()); return;
}
printf("%d\n", getdep(x)); cut(x, Fa[x]);
if(lson[x]) cut(x, lson[x]), link(lson[x], Fa[x]);
rson[Fa[x]] = lson[x]; if(lson[x]) Fa[lson[x]] = Fa[x];
set.erase(--set.end());
}
int main()
{
m = read();
for(RG int i = 1; i <= m; i++)
{
int opt = read();
switch(opt)
{
case 1: insert(read()); break;
case 2: splay_min(); break;
case 3: splay_max(); break;
case 4: del_min(); break;
case 5: del_max(); break;
}
}
return 0;
}