长链剖分

前言

某道题要求树上 (k) 级祖先,然后我发现我不会长剖,学了一下,感觉会了重剖之后长剖很好理解。

但是我为什么不会长剖呢?

讲解

先鸽着。

练习

板题(洛谷)

代码

板题(洛谷)
//12252024832524
#include <cstdio>
#include <cstring>
#include <algorithm>
#define TT template<typename T>
using namespace std; 

typedef long long LL;
const int MAXN = 500005;
int n,Q;

LL Read()
{
	LL x = 0,f = 1;char c = getchar();
	while(c > '9' || c < '0'){if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9'){x = (x*10) + (c^48);c = getchar();}
	return x * f;
}
TT void Put1(T x)
{
	if(x > 9) Put1(x/10);
	putchar(x%10^48);
}
TT void Put(T x,char c = -1)
{
	if(x < 0) putchar('-'),x = -x;
	Put1(x); if(c >= 0) putchar(c);
}
TT T Max(T x,T y){return x > y ? x : y;}
TT T Min(T x,T y){return x < y ? x : y;}
TT T Abs(T x){return x < 0 ? -x : x;}

#define ui unsigned int
ui S;

inline ui Get() {
	S ^= S << 13;
	S ^= S >> 17;
	S ^= S << 5;
	return S; 
}

int head[MAXN],tot;
struct edge
{
	int v,nxt;
}e[MAXN];
void Add_Edge(int x,int y)
{
	e[++tot].v = y;
	e[tot].nxt = head[x];
	head[x] = tot;
}

int f[MAXN][19],d[MAXN],LG[MAXN],rt,son[MAXN],MD[MAXN];
void dfs1(int x)
{
	MD[x] = d[x] = d[f[x][0]] + 1;
	for(int i = 1;i <= LG[d[x]];++ i) f[x][i] = f[f[x][i-1]][i-1];
	for(int i = head[x]; i ;i = e[i].nxt)
	{
		dfs1(e[i].v);
		MD[x] = Max(MD[x],MD[e[i].v]);
		if(MD[e[i].v] > MD[son[x]]) son[x] = e[i].v;
	}
}
int dfn[MAXN],dfntot,U[MAXN],D[MAXN],tp[MAXN];
void dfs2(int x,int t) 
{
	if(x == t) tp[x] = x;
	else tp[x] = tp[f[x][0]];
	dfn[x] = ++dfntot;
	U[dfntot] = t; D[dfntot] = x;
	if(!son[x]) return;
	dfs2(son[x],f[t][0]);
	for(int i = head[x]; i ;i = e[i].nxt)
		if(e[i].v^son[x])
			dfs2(e[i].v,e[i].v);
}
int Query(int u,int k)
{
	if(!k) return u;
	u = f[u][LG[k]]; k -= 1<<LG[k];
	k -= d[u] - d[tp[u]]; u = tp[u];
	if(k >= 0) return U[dfn[u]+k];
	else return D[dfn[u]-k];
}

int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	n = Read(); Q = Read(); S = Read();
	LG[0] = -1;
	for(int i = 1;i <= n;++ i) LG[i] = LG[i>>1] + 1;
	for(int i = 1;i <= n;++ i)
		if((f[i][0] = Read())) Add_Edge(f[i][0],i);
		else rt = i;
	dfs1(rt);
	dfs2(rt,rt);
	LL ans = 0,lst = 0;
	for(int i = 1;i <= Q;++ i)
	{
		int u = (Get()^lst) % n + 1;
		int k = (Get()^lst) % d[u];
		lst = Query(u,k);
		ans ^= lst * i;
	}
	Put(ans);
	return 0;
}
原文地址:https://www.cnblogs.com/PPLPPL/p/15234040.html