You ye Jiu yuan is the daughter of the Great GOD Emancipator. And when she becomes an adult, she will be queen of Tusikur, so she wanted to travel the world while she was still young. In a country, she found a small pub called Whitehouse. Just as she was about to go in for a drink, the boss Carola appeared. And ask her to solve this problem or she will not be allowed to enter the pub. The problem description is as follows:
There is a tree with
nn
n nodes, each node
ii
i contains weight
a[i]a[i]
a[i], the initial value of
a[i]a[i]
a[i] is
00
0. The root number of the tree is
11
1. Now you need to do the following operations:
1)1)
1) Multiply all weight on the path from
uu
u to
vv
v by
xx
x
2)2)
2) For all weight on the path from
uu
u to
vv
v, increasing
xx
x to them
3)3)
3) For all weight on the path from
uu
u to
vv
v, change them to the bitwise NOT of them
4)4)
4) Ask the sum of the weight on the path from
uu
u to
vv
v
The answer modulo
2642^{64}
2
64
.
Jiu Yuan is a clever girl, but she was not good at algorithm, so she hopes that you can help her solve this problem. Ding
∽∽∽\backsim\backsim\backsim
∽∽∽
The bitwise NOT is a unary operation that performs logical negation on each bit, forming the ones’ complement of the given binary value. Bits that are
00
0 become
11
1, and those that are
11
1 become
00
0. For example:
NOT 0111 (decimal 7) = 1000 (decimal 8)
NOT 10101011 = 01010100
Input
The input contains multiple groups of data.
For each group of data, the first line contains a number of
nn
n, and the number of nodes.
The second line contains
(n−1)(n - 1)
(n−1) integers
bib_i
b
i
, which means that the father node of node
(i+1)(i +1)
(i+1) is
bib_i
b
i
.
The third line contains one integer
mm
m, which means the number of operations，
The next
mm
m lines contain the following four operations:
At first, we input one integer opt
1)1)
1) If opt is
11
1, then input
33
3 integers,
u,v,xu, v, x
u,v,x, which means multiply all weight on the path from
uu
u to
vv
v by
xx
x
2)2)
2) If opt is
22
2, then input
33
3 integers,
u,v,xu, v, x
u,v,x, which means for all weight on the path from
uu
u to
vv
v, increasing
xx
x to them
3)3)
3) If opt is
33
3, then input
22
2 integers,
u,vu, v
u,v, which means for all weight on the path from
uu
u to
vv
v, change them to the bitwise NOT of them
4)4)
4) If opt is
44
4, then input
22
2 integers,
u,vu, v
u,v, and ask the sum of the weights on the path from
uu
u to
vv
v
1≤n,m,u,v≤1051 \le n,m,u,v \le 10^5
1≤n,m,u,v≤10
5
1≤x<2641 \le x < 2^{64}
1≤x<2
64
Output
For each operation
44
4, output the answer.
样例输入
复制
7
1 1 1 2 2 4
5
2 5 6 1
1 1 6 2
4 5 6
3 5 2
4 2 2
2
1
4
3 1 2
4 1 2
3 1 1
4 1 1
样例输出
复制
5
18446744073709551613
18446744073709551614
0
一看就是树链剖分，只不过当时不知道位反怎么写，后来看了别人的博客才发现很简单的退一下就好了，只是我题目理解错了。位反就是
，这个-x怎么算呢，因为他是mod
就是unsigned long long 的自动取模，所以-x就变为
，那么就变成
了，就可以用加和乘来存，有一点显而易见的知识要知道：val*(x+B)=val*x+val*B，所以addf数组也要乘上val。
#include<bits/stdc++.h>
using namespace std;
#define ll unsigned long long
ll inf=18446744073709551615;
const int maxn=2e5+5;
struct node
{
int to,next;
}e[maxn*2];
int head[maxn],cnt,tim,fa[maxn],son[maxn],dep[maxn],top[maxn],en[maxn],siz[maxn],pos[maxn];
ll mulf[maxn*4],addf[maxn*4],sum[maxn*4],a[maxn];
int n,m,rt;
void add(int x,int y)
{
e[cnt].to=y;
e[cnt].next=head[x];
head[x]=cnt++;
}
void dfs1(int u)
{
siz[u]=1;
for (int i=head[u];~i;i=e[i].next)
{
int v=e[i].to;
if (v!=fa[u])
{
fa[v]=u;
dep[v]=dep[u]+1;
dfs1(v);
siz[u]+=siz[v];
if (son[u]==0||siz[v]>siz[son[u]])
son[u]=v;
}
}
}
void dfs2(int u,int f)
{
tim++;
top[u]=f;
pos[u]=tim;
if (son[u]) dfs2(son[u],f);
for (int i=head[u];~i;i=e[i].next)
{
int v=e[i].to;
if (v!=fa[u]&&v!=son[u])
{
dfs2(v,v);
}
}
en[u]=tim;
}
void pushup(int root)
{
sum[root]=sum[root<<1]+sum[root<<1|1];
}
void pushdown(int l,int r,int root)
{
int mid=l+r>>1;
sum[root<<1]=sum[root<<1]*mulf[root]+addf[root]*(mid-l+1);
sum[root<<1|1]=sum[root<<1|1]*mulf[root]+addf[root]*(r-mid);
mulf[root<<1]=mulf[root<<1]*mulf[root];
mulf[root<<1|1]=mulf[root<<1|1]*mulf[root];
addf[root<<1]=addf[root<<1]*mulf[root]+addf[root];
addf[root<<1|1]=addf[root<<1|1]*mulf[root]+addf[root];
addf[root]=0;
mulf[root]=1;
}
void add(int l,int r,int root,int ql,int qr,ll val,int op)
{
if(l>=ql&&r<=qr)
{
if(op==1)
{
mulf[root]*=val;
sum[root]*=val;
addf[root]*=val;
}
else if(op==2)
{
sum[root]+=(r-l+1)*val;
addf[root]+=val;
}
else
{
sum[root]=inf*sum[root]+(r-l+1)*inf;
addf[root]=inf*addf[root]+inf;
mulf[root]*=inf;
}
return ;
}
pushdown(l,r,root);
int mid=l+r>>1;
if(mid>=ql)
add(l,mid,root<<1,ql,qr,val,op);
if(mid<qr)
add(mid+1,r,root<<1|1,ql,qr,val,op);
pushup(root);
}
ll ask(int l,int r,int root,int ql,int qr)
{
if(l>=ql&&r<=qr)
return sum[root];
int mid=l+r>>1;
pushdown(l,r,root);
ll ans=0;
if(mid>=ql)
ans+=ask(l,mid,root<<1,ql,qr);
if(mid<qr)
ans+=ask(mid+1,r,root<<1|1,ql,qr);
return ans;
}
void update(int x,int y,ll val,int op)
{
while (top[x]!=top[y])
{
if (dep[top[x]]<dep[top[y]]) swap(x,y);
add(1,n,1,pos[top[x]],pos[x],val,op);
x=fa[top[x]];
}
if (pos[x]>pos[y]) swap(x,y);
add(1,n,1,pos[x],pos[y],val,op);
}
void query(int x,int y)
{
ll ans=0;
while (top[x]!=top[y])
{
if (dep[top[x]]<dep[top[y]]) swap(x,y);
ans+=ask(1,n,1,pos[top[x]],pos[x]);
x=fa[top[x]];
}
if (pos[x]>pos[y]) swap(x,y);
ans+=ask(1,n,1,pos[x],pos[y]);
printf("%llu\n",ans);
}
int main()
{
while(~scanf("%d",&n))
{
memset(head,-1,sizeof(head));
for(int i=0;i<maxn;i++)
fa[i]=son[i]=dep[i]=top[i]=siz[i]=pos[i]=0;
cnt=tim=0;
for(int i=0;i<maxn*4;i++)
mulf[i]=1,addf[i]=sum[i]=0;
int sta,fin;
ll val;
for(int i=2;i<=n;i++)
{
scanf("%d",&sta);
add(sta,i),add(i,sta);
}
rt=1;
dfs1(rt);
dfs2(rt,rt);
int op;
scanf("%d",&m);
for(int i=1;i<=m;i++)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d%llu",&sta,&fin,&val);
update(sta,fin,val,op);
}
else if(op==2)
{
scanf("%d%d%llu",&sta,&fin,&val);
update(sta,fin,val,op);
}
else if(op==3)
{
scanf("%d%d",&sta,&fin);
update(sta,fin,0,op);
}
else
{
scanf("%d%d",&sta,&fin);
query(sta,fin);
}
}
}
return 0;
}