题目描述
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 n nodes, each node i contains weight a[i], the initial value of a[i] is 0. The root number of the tree is 1. Now you need to do the following operations:
1) Multiply all weight on the path from u to v by x
2) For all weight on the path from u to v, increasing x to them
3) For all weight on the path from u to v, change them to the bitwise NOT of them
4) Ask the sum of the weight on the path from u to v
The answer modulo 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~~~
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 0 become 1, and those that are 1 become 0. For example:
NOT 0111 (decimal 7) = 1000 (decimal 8)
NOT 10101011 = 01010100
输入
The input contains multiple groups of data.
For each group of data, the first line contains a number of n, and the number of nodes.
The second line contains (n - 1) integers bi, which means that the father node of node (i +1) is bi.
The third line contains one integer m, which means the number of operations，
The next m lines contain the following four operations:
At first, we input one integer opt
1) If opt is 1, then input 3 integers, u, v, x, which means multiply all weight on the path from u to v by x
2) If opt is 2, then input 3 integers, u, v, x, which means for all weight on the path from u to v, increasing x to them
3) If opt is 3, then input 2 integers, u, v, which means for all weight on the path from u to v, change them to the bitwise NOT of them
4) If opt is 4, then input 2 integers, u, v, and ask the sum of the weights on the path from u to v
1 ≤ n, m, u, v ≤ 10^5
1 ≤ x < 2^64
输出
For each operation 4, output the answer.
样例输入
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
5 18446744073709551613 18446744073709551614 0
题意 给一棵n个节点的有根树，每个节点有权值，初始是0，m次操作 1 u v x：给u v路径上的点权值*x 2 u v x：给u v路径上的点权值+x 3 u v：给u v路径上的点权值取反 4 u v：询问u v路径上的权值和，对2^64取模 树链剖分：https://wenku.baidu.com/view/a088de01eff9aef8941e06c3.html 然后如果没有取反操作，线段树维护和sum，加法标记add和乘法标记mul即可 对于取反操作，因为是对2^64取模的，即x+(!x)=2^64-1，所以x=(2^64-1)-x，因此取反就变成乘法和加法了：!x=(-1)*x+(-1) （-1对于2^64取模后是（2^64-1)）
#include <bits/stdc++.h>
#define ull unsigned long long
using namespace std;
const int N=1e5+100;
int n,m,tot,cnt;
int fa[N],last[N];
int son[N],deep[N],dfn[N],num[N],top[N];//重儿子 深度 dfs序 子树规模 所在重链的顶端节点
ull sum[N*4],add[N*4],mul[N*4];
struct orz{
int v,nex;}e[N];
void init()
{
cnt=0;
tot=0;
memset(last,0,sizeof(last));
memset(son,-1,sizeof(son));
}
void Inses(int x,int y)
{
cnt++;
e[cnt].v=y;
e[cnt].nex=last[x];
last[x]=cnt;
}
void dfs1(int x,int d)
{
deep[x]=d;
num[x]=1;
for (int i=last[x];i;i=e[i].nex)
{
int v=e[i].v;
dfs1(v,d+1);
num[x]+=num[v];
if (son[x]==-1 || num[v]>num[son[x]]) son[x]=v;
}
}
void dfs2(int x,int sp)
{
top[x]=sp;
dfn[x]=++tot;
if (son[x]==-1) return ;
dfs2(son[x],sp);
for (int i=last[x];i;i=e[i].nex)
{
int v=e[i].v;
if (v!=son[x]) dfs2(v,v);
}
}
void PushUp(int s)
{
sum[s]=sum[s<<1]+sum[s<<1|1];
}
void PushDown(int s,int l,int r)
{
if (mul[s]!=1)
{
mul[s<<1]*=mul[s];
mul[s<<1|1]*=mul[s];
add[s<<1]*=mul[s];
add[s<<1|1]*=mul[s];
sum[s<<1]*=mul[s];
sum[s<<1|1]*=mul[s];
mul[s]=1;
}
if (add[s])
{
add[s<<1]+=add[s];
add[s<<1|1]+=add[s];
int mid=(l+r)>>1;
sum[s<<1]+=(ull)(mid-l+1)*add[s];
sum[s<<1|1]+=(ull)(r-mid)*add[s];
add[s]=0;
}
}
void build(int s,int l,int r)
{
sum[s]=add[s]=0;
mul[s]=1;
if (l==r) return ;
int m=(l+r)>>1;
build(s<<1,l,m); build(s<<1|1,m+1,r);
PushUp(s);
}
void update(int s,int l,int r,int L,int R,ull val,int op)
{
//printf("s=%d,l=%d,r=%d,L=%d,R=%d\n",s,l,r,L,R);
if (L<=l&&r<=R)
{
if (l!=r) PushDown(s,l,r);
if (op==1)
{
mul[s]*=val;
add[s]*=val;
sum[s]*=val;
}
else if (op==2)
{
add[s]+=val;
sum[s]+=(ull)(r-l+1)*val;
}
else
{
mul[s]*=val;
add[s]*=val;
add[s]+=val;
sum[s]=(ull)(r-l+1)*val-sum[s];
}
return;
}
PushDown(s,l,r);
int mid=(l+r)>>1;
if (L<=mid) update(s<<1,l,mid,L,R,val,op);
if (R>mid) update(s<<1|1,mid+1,r,L,R,val,op);
PushUp(s);
}
ull query(int s,int l,int r,int L,int R)
{
if (L<=l&&r<=R) return sum[s];
PushDown(s,l,r);
int mid=(l+r)>>1;
ull ans=0;
if (L<=mid) ans+=query(s<<1,l,mid,L,R);
if (R>mid) ans+=query(s<<1|1,mid+1,r,L,R);
PushUp(s);
return ans;
}
void solve(int op,int x, int y,ull val)
{
if (op==3) val=-1;
if (op<=3)
{
while (top[x]!=top[y])
{
if (deep[top[x]]<deep[top[y]]) swap(x, y);
update(1,1,n,dfn[top[x]],dfn[x],val,op);
x=fa[top[x]];
}
if (deep[x]>deep[y]) swap(x,y);
update(1,1,n,dfn[x],dfn[y],val,op);
}
else
{
ull ans=0;
while (top[x]!=top[y])
{
if (deep[top[x]]<deep[top[y]]) swap(x, y);
ans+=query(1,1,n,dfn[top[x]],dfn[x]);
x=fa[top[x]];
}
if (deep[x]>deep[y]) swap(x,y);
ans+=query(1,1,n,dfn[x],dfn[y]);
printf("%llu\n",ans);
}
}
int main()
{
while (scanf("%d",&n)!=EOF)
{
init();
for (int i=2;i<=n;i++)
{
scanf("%d",&fa[i]);
Inses(fa[i],i);
}
dfs1(1,0);
dfs2(1,1);
build(1,1,n);
scanf("%d",&m);
int op,u,v; ull x;
while (m--)
{
scanf("%d",&op);
if (op==1 || op==2) scanf("%d%d%llu",&u,&v,&x);
else scanf("%d%d",&u,&v);
solve(op,u,v,x);
}
}
return 0;
}
原文地址：https://www.cnblogs.com/tetew/p/11293766.html