Jiu Yuan Wants to Eat

Posted 2022-06-09 czy-power

问题 E: Jiu Yuan Wants to Eat

时间限制: 4 Sec  内存限制: 128 MB
提交: 89  解决: 35
[提交] [状态] [命题人:admin]

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

复制样例数据

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

 

#include <bits/stdc++.h>

#define ll unsigned long long
using namespace std;
//typedef long long ll;
const int maxn = 300600;
//const ll mod = 1000000007;

const ll mod = 18446744073709551615;
struct node 
    int to, nx;
 p[maxn];
struct segment_tree 
    int l, r;
    ll val, add_tag;
    ll mul_tag;
 s[maxn];
int n, q, tot;
int head[maxn];
int fa[maxn], dep[maxn];
int siz[maxn];
int son[maxn];
int top[maxn], dfn[maxn], id[maxn];

void add_edge(int u, int v) 
    p[++tot].to = v;
    p[tot].nx = head[u];
    head[u] = tot;


void dfs1(int u) 
    siz[u] = 1;
    for (int i = head[u]; i; i = p[i].nx) 
        int to = p[i].to;
        if (to == fa[u])continue;
        fa[to] = u;
        dep[to] = dep[u] + 1;
        dfs1(to);
        siz[u] += siz[to];
        if (siz[to] > siz[son[u]]) 
            son[u] = to;
        
    


void dfs2(int u, int tp) 
    top[u] = tp;
    dfn[u] = ++tot;
    id[tot] = u;
    if (!son[u])return;
    dfs2(son[u], tp);
    for (int i = head[u]; i; i = p[i].nx) 
        int to = p[i].to;
        if (to != fa[u] && to != son[u]) 
            dfs2(to, to);
        
    


void push_up(int rt) 
    s[rt].val = s[rt << 1].val + s[rt << 1 | 1].val;


void build(int l, int r, int rt) 
    s[rt].l = l;
    s[rt].r = r;
    s[rt].val = 0;
    s[rt].add_tag = 0;
    s[rt].mul_tag = 1;
    if (l == r)return;
    int mid = l + r >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    push_up(rt);


void pushdown(int rt) 
    if (s[rt].add_tag != 0 || s[rt].mul_tag != 1) 
        s[rt << 1].val *= s[rt].mul_tag;
        s[rt << 1].val += (s[rt].add_tag * (s[rt << 1].r - s[rt<<1].l + 1));
        s[rt << 1].add_tag = s[rt << 1].add_tag * s[rt].mul_tag + s[rt].add_tag;
        s[rt << 1].mul_tag *= s[rt].mul_tag;
        s[rt << 1 | 1].val *= s[rt].mul_tag;
        s[rt << 1 | 1].val += (s[rt].add_tag * (s[rt << 1|1].r - s[rt << 1|1].l + 1));
        s[rt << 1 | 1].add_tag = s[rt << 1 | 1].add_tag * s[rt].mul_tag + s[rt].add_tag;
        s[rt << 1 | 1].mul_tag *= s[rt].mul_tag;
        s[rt].add_tag = 0;
        s[rt].mul_tag = 1;
    


void upd(int op, int ql, int qr, ll w, int rt) 
    if (ql <= s[rt].l && qr >= s[rt].r) 
        if (op == 1) 
            s[rt].val += w * (s[rt].r - s[rt].l + 1);
            s[rt].add_tag += w;
         else if (op == 2) 
            s[rt].val *= w;
            s[rt].add_tag *= w;
            s[rt].mul_tag *= w;
        
        return;
    
    pushdown(rt);
    if (ql <= s[rt << 1].r) 
        upd(op, ql, qr, w, rt << 1);
    
    if (qr >= s[rt << 1 | 1].l) 
        upd(op, ql, qr, w, rt << 1 | 1);
    
    push_up(rt);


void update(int op, int x, int y, ll w) 
    int fx = top[x], fy = top[y];
    while (fx != fy) 
        if (dep[fx] <= dep[fy]) 
            upd(op, dfn[fy], dfn[y], w, 1);
            y = fa[fy];
            fy = top[y];
         else 
            upd(op, dfn[fx], dfn[x], w, 1);
            x = fa[fx];
            fx = top[x];
        
    
    if (dep[x] <= dep[y]) 
        upd(op, dfn[x], dfn[y], w, 1);
     else 
        upd(op, dfn[y], dfn[x], w, 1);
    


ll Query(int ql, int qr, int rt) 
    if (ql <= s[rt].l && qr >= s[rt].r) 
        return s[rt].val;
    
    pushdown(rt);
    ll ans = 0;
    if (ql <= s[rt << 1].r) 
        ans += Query(ql, qr, rt << 1);
    
    if (qr >= s[rt << 1 | 1].l) 
        ans += Query(ql, qr, rt << 1 | 1);
    
    return ans;


ll query(int x, int y) 
    int fx = top[x], fy = top[y];
    ll ans = 0;
    while (fx != fy) 
        if (dep[fx] <= dep[fy]) 
            ans += Query(dfn[fy], dfn[y], 1);
            y = fa[fy];
            fy = top[y];
         else 
            ans += Query(dfn[fx], dfn[x], 1);
            x = fa[fx];
            fx = top[x];
        
    
    if (dep[x] <= dep[y]) 
        ans += Query(dfn[x], dfn[y], 1);
     else 
        ans += Query(dfn[y], dfn[x], 1);
    
    return ans;


int main() 
    //freopen("1.txt", "r", stdin);
    while (~scanf("%d", &n)) 
        memset(head, 0, sizeof(head));
        tot = 0;
        memset(top, 0, sizeof(top));
        memset(id, 0, sizeof(id));
        memset(son,0,sizeof(son));
        memset(dep,0,sizeof(dep));
        for (int i = 2, u; i <= n; ++i) 
            scanf("%d", &u);
            add_edge(i, u);
            add_edge(u, i);
        
        fa[1] = 0;
        dep[1] = 1;
        dfs1(1);
        tot = 0;
        dfs2(1, 1);

        build(1, n, 1);

        scanf("%d", &q);
        int op, u, v;
        ll x;
//        for(int i=1;i<=n;++i)printf("%d ",top[i]);
//        cout<<endl;
//        for(int i=1;i<=n;++i)printf("%d ",dep[i]);
//        cout<<endl;
//        for(int i=1;i<=n;++i)printf("%d ",dfn[i]);
//        cout<<endl;
        while (q--) 
            scanf("%d", &op);
            if (op == 1) 
                scanf("%d%d%llu", &u, &v, &x);

                update(2, u, v, x);

             else if (op == 2) 
                scanf("%d%d%llu", &u, &v, &x);
                //cout<<"&*"<<endl;
                update(1, u, v, x);
                //cout<<"debug op="<<op<<endl;
             else if (op == 3) 
                scanf("%d%d", &u, &v);
                //cout<<"&*"<<endl;
                update(2, u, v, -1);
                update(1, u, v, mod);
                //cout<<"last&*"<<endl;
             else 
                scanf("%d%d", &u, &v);
                printf("%llu\n", query(u, v));
            
        
    
    return 0;

/*
1
5 3 -3
-1 -2 3 -4 -5
-+*
*/