2017 ACM西安网络赛 G题 Xor 树链剖分+分块

There is a tree with nn nodes. For each node, there is an integer value a_iai, (1 \le a_i \le 1,000,000,0001≤ai≤1,000,000,000 for 1 \le i \le n1≤i≤n). There is qq queries which are described as follow: Assume the value on the path from node aa to node bb is t_0, t_1, \cdots t_mt0,t1,⋯tm. You are supposed to calculate t_0t0 xor t_ktk xor t_{2k}t2k xor ... xor t_{pk}tpk (pk \le m)(pk≤m).

Input Format

There are multi datasets. (\sum n \le 50,000, \sum q \le 500,000)(∑n≤50,000,∑q≤500,000).

For each dataset: In the first n-1n−1 lines, there are two integers u,vu,v, indicates there is an edge connect node uuand node vv.

In the next nn lines, There is an integer a_iai (1 \le a_i \le 1,000,000,0001≤ai≤1,000,000,000).

In the next qq lines, There is three integers a,ba,b and kk. (1 \le a,b,k \le n1≤a,b,k≤n).

Output Format

For each query, output an integer in one line, without any additional space.

样例输入复制

5 6 1 5 4 1 2 1 3 2 19 26 0 8 17 5 5 1 1 3 2 3 2 1 5 4 2 3 4 4 1 4 5

样例输出复制

17 19 26 25 0 19

题目来源

​​2017 ACM-ICPC 亚洲区（西安赛区）网络赛​​

题意：

给出一棵树，每个节点有个val，q个询问。 
询问格式：u, v, k, 从u到v的路径， 每个k个点取一个，求所有点的val值异或和。

分析：

BZOJ修改版本，具体解析：​​BZOJ4381: [POI2015]Odwiedziny 分块+树链剖分​​

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
int read()
{
    int x=0,f=1;
    char ch=getchar();
    while (ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while (ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
typedef unsigned long long ULL;

const ULL INF=(ULL)0-(ULL)1;

const int N = 1e5+5;
int a[N],b[N],c[N];
int up[N][235],block;//up[x,y]表示从点x开始每次跳y步直到不能跳为止的权值和
int n;
int w[N];
int head[N], cnt;
int num;
int root[N];//保存结点 u 的父亲节点
int depth[N]; //保存结点 u 的深度值
int W_son[N];//保存重儿子
int size[N]; //保存以 u 为根的子树节点个数
int top[N];//保存当前节点所在链的顶端节点
int id[N];//保存树中每个节点剖分以后的新编号（ DFS 的执行顺序）
int new_W[N]; //保存当前 dfs 标号在树中所对应的节点
int dfn[N];// 保存dfs序，dfn[i]表示dfs序号为i的原数组下标

struct Eddge
{
    int nex, to;
} edge[N<<1];//这里卡了好久，一定要乘以2
void addEddge(int u, int v)
{

    edge[++cnt].to=v;
    edge[cnt].nex=head[u];
    head[u] = cnt;
}
void dfs1(int u, int fa, int deep)
{

    depth[u] = deep;
    root[u] = fa;
    size[u] = 1;
    int v=u;
    for (int i=1; i<=block; i++)
        v=root[v],up[u][i]=a[u]^up[v][i];

    int maxSon = -1;
    for(int i=head[u]; i!=-1; i=edge[i].nex)
    {
        int v = edge[i].to;
        if(v == fa)
            continue;
        dfs1(v, u, deep+1);
        size[u] += size[v];
        if(size[v] > maxSon)
        {
            maxSon = size[v];
            W_son[u] = v;
        }
    }
}
void dfs2(int u, int topf)
{

    id[u] = ++num;
    dfn[num]=u;
    new_W[num] = a[u];
    top[u] = topf;
    if(!W_son[u])
        return;
    dfs2(W_son[u], topf);
    for(int i=head[u]; i!=-1; i=edge[i].nex)
    {
        int v = edge[i].to;
        if(v == W_son[u] || v == root[u])
            continue;
        dfs2(v, v);
    }
}

int get_lca(int x,int y)
{
    while (top[x]!=top[y])
    {
        if (depth[top[x]]<depth[top[y]])
            swap(x,y);
        x=root[top[x]];
    }
    return depth[x]<depth[y]?x:y;
}
//获取x的k级祖先
int get_kth(int x,int k)
{
    while (x&&depth[x]-depth[top[x]]<k)
        k-=depth[x]-depth[top[x]]+1,x=root[top[x]];
    return !x?0:dfn[id[x]-k];
}
//返回x到y，步长为z的异或和
int get_sum(int x,int y,int z)
{
    int lca=get_lca(x,y);
    if (z<=block)
    {
        int ans=up[x][z]^up[get_kth(lca,z-(depth[x]-depth[lca])%z)][z];
        int tmp=get_kth(y,(depth[x]+depth[y]-depth[lca]*2)%z);
        if (depth[tmp]>=depth[lca])
            ans^=up[tmp][z]^up[get_kth(lca,z-(depth[tmp]-depth[lca])%z)][z];
        /*if ((depth[x]+depth[y]-depth[lca]*2)%z>0)///x到y的步数无法整除z，敲好不能到终点y，
           ans^=a[y]; */                           ///终点y的值计算在内，则消除注释
            
        if ((depth[x]-depth[lca])%z==0)
            ans^=a[lca];
         return ans;
    }
    else
    {
        int ans=0,tmp=x;
        while (depth[tmp]>=depth[lca])
            ans^=a[tmp],tmp=get_kth(tmp,z);
        tmp=get_kth(y,(depth[x]+depth[y]-depth[lca]*2)%z);
        while (depth[tmp]>=depth[lca])
            ans^=a[tmp],tmp=get_kth(tmp,z);

        /*if ((depth[x]+depth[y]-depth[lca]*2)%z>0)///x到y的步数无法整除z，敲好不能到终点y，
            ans^=a[y];   */ ///终点y的值计算在内，则消除注释
        if ((depth[x]-depth[lca])%z==0)
            ans^=a[lca];
        return ans;
    }
}

void init()
{
    cnt = num = 0;
    memset(head, -1, sizeof(head));
    memset(W_son, 0, sizeof(W_son));
}

int main()
{

    while(scanf("%d",&n)!=-1)
    {
        init();
        int m;
        m=read();
        for (int i=1; i<n; i++)
        {
            int x=read(),y=read();
            addEddge(x,y);
            addEddge(y,x);
        }
        block=sqrt(n+0.5);
        for (int i=1; i<=n; i++)
            a[i]=read();

        dfs1(1,0,1);
        dfs2(1,1);
        while(m--)
        {
            int x,y,z;
            x=read();
            y=read();
            z=read();
            printf("%d\n",get_sum(x,y,z));
        }
    }


    return 0;
}