焦作网络赛E-JiuYuanWantstoEat【树链剖分】【线段树】

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 nodes, each node ii contains weight a[i]a[i], the initial value of a[i]a[i] is 00. The root number of the tree is 11. Now you need to do the following operations:

1)1) Multiply all weight on the path from uu to vv by xx

2)2) For all weight on the path from uu to vv, increasing xx to them

3)3) For all weight on the path from uu to vv, change them to the bitwise NOT of them

4)4) Ask the sum of the weight on the path from uu to vv

The answer modulo 2^{64}264.

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 become 11, and those that are 11 become 00. 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, and the number of nodes.

The second line contains (n - 1)(n−1) integers b_ibi​, which means that the father node of node (i +1)(i+1) is b_ibi​.

The third line contains one integer mm, which means the number of operations，

The next mm lines contain the following four operations:

At first, we input one integer opt

1)1) If opt is 11, then input 33 integers, u, v, xu,v,x, which means multiply all weight on the path from uu to vv by xx

2)2) If opt is 22, then input 33 integers, u, v, xu,v,x, which means for all weight on the path from uu to vv, increasing xx to them

3)3) If opt is 33, then input 22 integers, u, vu,v, which means for all weight on the path from uu to vv, change them to the bitwise NOT of them

4)4) If opt is 44, then input 22 integers, u, vu,v, and ask the sum of the weights on the path from uu to vv

1 \le n,m,u,v \le 10^51≤n,m,u,v≤105

1 \le x < 2^{64}1≤x<264

Output

For each operation 44, 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

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

 

题意：

有一棵树 4种类型的操作

1 u v x表示将u到v路径上的点的值乘以x

2 u v x表示将u到v路径上的点的值加x

3 u v 表示将u到v路径上的点的值取反

4 u v 表示查询u到v路径上所有点值之和

答案取模2^64

思路：

虽然操作乍一看就是线段树 但是和路径相关需要用的树链剖分了

124都是常见操作 只有3比较麻烦

应该要考虑到（-x）%(2^64) = (2^64-1)*x%(2^64)

-x = !x + 1

！x = (2^64-1)*x + (2^64-1) 就可以转换为乘一个数再加一个数了

因此线段树用三个数组维护 一个是sum存区间之和 add是加的lazy数组 mul是乘的lazy数组

由于答案取模2^64 比较特殊

用unsigned long long 位数刚好 溢出相当于取模

由于用到了dfs序 写的时候要注意标号的起始

//#include"pch.h"

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
#include<cstring>
#include<set>
#include<stack>
//#include<bits/stdc++.h>

#define inf 18446744073709551615
using namespace std;
typedef unsigned long long LL;

const int MAXN = 2e5 + 5;
int siz[MAXN];//number of son
int top[MAXN];//top of the heavy link
int son[MAXN];//heavy son of the node
int dep[MAXN];//depth of the node
int faz[MAXN];//father of the node
int tid[MAXN];//ID -> DFSID
int rnk[MAXN];//DFSID -> ID
int head[MAXN], cnt, n, m, cntid;
LL sum[MAXN << 2], add[MAXN << 2], mul[MAXN << 2];
struct edge {
    int to;
    int next;
}edg[MAXN];

void addedge(int u, int v)
{
    edg[cnt].to = v;
    edg[cnt].next = head[u];
    head[u] = cnt++;
}

void dfs1(int u, int father, int depth)
{
    dep[u] = depth;
    faz[u] = father;
    siz[u] = 1;

    for (int i = head[u]; i != -1; i = edg[i].next) {
        int v = edg[i].to;
        if (v != faz[u]) {
            dfs1(v, u, depth + 1);
            siz[u] += siz[v];
            if (son[u] == -1 || siz[v] > siz[son[u]]) {
                son[u] = v;
            }
        }
    }
}

void dfs2(int u, int t)
{
    top[u] = t;
    tid[u] = cntid;
    rnk[cntid] = u;
    cntid++;

    if (son[u] == -1) {
        return;
    }
    dfs2(son[u], t);
    for (int i = head[u]; i != -1; i = edg[i].next) {
        int v = edg[i].to;
        if (v != son[u] && v != faz[u]) {
            dfs2(v, v);
        }
    }
}

void pushup(int rt)
{
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}

void pushdown(int rt, int l, int r)
{
    add[rt << 1] = add[rt << 1] * mul[rt] + add[rt];
    add[rt << 1 | 1] = add[rt << 1 | 1] * mul[rt] + add[rt];
    mul[rt << 1] = mul[rt << 1] * mul[rt];
    mul[rt << 1 | 1] = mul[rt << 1 | 1] * mul[rt];
    int m = (l + r) / 2;
    sum[rt << 1] = sum[rt << 1] * mul[rt] + add[rt] * (m - l + 1);
    sum[rt << 1 | 1] = sum[rt << 1 | 1] * mul[rt] + add[rt] * (r - m);
    add[rt] = 0;
    mul[rt] = 1;
}

void build(int rt, int l, int r)
{
    if (l == r) {
        return;
    }
    int m = (l + r) / 2;
    build(rt << 1, l, m);
    build(rt << 1 | 1, m + 1, r);
    pushup(rt);
}

void update(int L, int R, LL c, int type, int l, int r, int rt)
{
    if (L <= l && R >= r) {
        if (type == 1) {
            sum[rt] = sum[rt] * c;
            add[rt] = add[rt] * c;
            mul[rt] = mul[rt] * c;
        }
        else if (type == 2) {
            sum[rt] = sum[rt] + (LL)c * (r - l + 1);
            add[rt] += c;
        }
        else if (type == 3) {
            sum[rt] = sum[rt] * inf + inf * (r - l + 1);
            add[rt] = add[rt] * inf + inf;
            mul[rt] *= inf;
        }
        return;
    }
    pushdown(rt, l, r);
    int m = (l + r) / 2;
    if (L <= m) {
        update(L, R, c, type, l, m, rt << 1);
    }
    if (R > m) {
        update(L, R, c, type, m + 1, r, rt << 1 | 1);
    }
    pushup(rt);
}

LL query(int L, int R, int l, int r, int rt)
{
    if (L <= l && R >= r) {
        return sum[rt];
    }
    int m = (l + r) / 2;
    LL res = 0;
    pushdown(rt, l, r);
    if (L <= m) {
        res += query(L, R, l, m, rt << 1);
    }
    if (R > m) {
        res += query(L, R, m + 1, r, rt << 1 | 1);
    }
    return res;
}

LL query_path(int x, int y)
{
    LL ans = 0;
    int fx = top[x], fy = top[y];
    while (fx != fy) {
        if (dep[fx] < dep[fy]) {
            swap(fx, fy);
            swap(x, y);
        }
        ans += query(tid[fx], tid[x], 1, n, 1);
        x = faz[fx];
        fx = top[x];
    }

    ans += (dep[x] > dep[y])?query(tid[y], tid[x], 1, n, 1):query(tid[x], tid[y], 1, n, 1);
    return ans;
}

void update_path(int x, int y, LL c, int type)
{
    int fx = top[x], fy = top[y];
    while (fx != fy) {
        if (dep[fx] < dep[fy]) {
            swap(fx, fy);
            swap(x, y);
        }
        update(tid[fx], tid[x], c, type, 1, n, 1);
        x = faz[fx];
        fx = top[x];
    }
    if(dep[x] < dep[y]){
        swap(x, y);
    }
    update(tid[y], tid[x], c, type, 1, n, 1);
}

void init()
{
    memset(head, -1, sizeof(head));
    memset(son, -1, sizeof(son));
    cnt = 0;
    cntid = 1;
    memset(add, 0, sizeof(add));
    memset(mul, 1, sizeof(mul));
    memset(sum, 0, sizeof(sum));
}

int main()
{
    while (scanf("%d", &n) != EOF) {
        init();
        for (int i = 1; i < n; i++) {
            int b;
            scanf("%d", &b);
            addedge(b, i + 1);
        }
        dfs1(1, 0, 1);
        dfs2(1, 1);
        build(1, 1, n);
        scanf("%d", &m);
        for (int i = 0; i < m; i++) {
            int op, u, v;
            LL x;
            scanf("%d%d%d", &op, &u, &v);
            if (op == 1 || op == 2) {
                scanf("%lld", &x);
            }
            if (op == 4) {
                printf("%llu\n", query_path(u, v));
            }
            else {
                if(op == 3){
                    update_path(u, v, 0, op);
                }
                else {
                    update_path(u, v, x, op);
                }
            }
        }
    }
    return 0;
}