2017 ACM-ICPC 亚洲区（西安赛区）网络赛 xor （根号分治）

xor

There is a tree with nn nodes. For each node, there is an integer value a_ia​i​​, (1 le a_i le 1,000,000,0001≤a​i​​≤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_mt​0​​,t​1​​,⋯t​m​​. You are supposed to calculate t_0t​0​​ xor t_kt​k​​ xor t_{2k}t​2k​​ xor ... xor t_{pk}t​pk​​ (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_ia​i​​ (1 le a_i le 1,000,000,0001≤a​i​​≤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

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

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】给您一棵树，每个节点有一个权值，q次询问，每次给出u,v,k，求从u到v的路径中，从u开始，每隔k个节点亦或一下的结果。

【分析】根号分治，大于根号n的暴力跳（倍增跳），小于根号n的用数组存起来，复杂度最高为O(N)*sqrt(N)*log(N).

  

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define rep(i,l,r) for(int i=(l);i<=(r);++i)
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 5e4+50;;
const int M = 255;
const int mod = 19260817;
const int mo=123;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,q,sz;
int a[N],fa[N][25],up[N][M+10],dep[N];
vector<int>edg[N],vec;
int find(int u,int k){
    for(int i=19;i>=0;i--){
        if(k>>i&1){
            u=fa[u][i];
            if(u==0)return 0;
        }
    }
    return u;
}
void dfs(int u,int f){
    fa[u][0]=f;
    for(int i=1;i<20;i++){
        fa[u][i]=fa[fa[u][i-1]][i-1];
    }
    for(int i=1;i<=sz;i++){
        up[u][i]=a[u];
        int v=find(u,i);
        up[u][i]^=up[v][i];
    }
    for(int i=0;i<edg[u].size();i++){
        int v=edg[u][i];
        if(v==f)continue;
        dep[v]=dep[u]+1;
        dfs(v,u);
    }
}
int LCA(int u,int v){
    int U=u,V=v;
    if(dep[u]<dep[v])swap(u,v);
    for(int i=19;i>=0;i--){
        if(dep[fa[u][i]]>=dep[v]){
            u=fa[u][i];
        }
    }
    if(u==v)return (u);
    for(int i=19;i>=0;i--){
        if(fa[u][i]!=fa[v][i]){
            u=fa[u][i];v=fa[v][i];
        }
    }
    return (fa[u][0]);
}
int solve(int u,int v,int k,int lca){
    int res=(dep[u]+dep[v]-2*dep[lca])%k;
    int U=u,V=v;
    v=find(v,res);
    int ans=0;
    while(dep[u]>=dep[lca]){
        ans^=a[u];
        u=find(u,k);
        if(!u)break;
    }
    if(V==lca||dep[v]<=dep[lca]||v==0)return ans;
    V=v;
    while(dep[v]>=dep[lca]){
        ans^=a[v];
        v=find(v,k);
        if(!v)break;
    }
    if((dep[U]-dep[lca])%k==0&&(dep[V]-dep[lca])%k==0)ans^=a[lca];
    return ans;
}
void init(){
    met(fa,0);met(up,0);
    for(int i=0;i<N;i++){
        edg[i].clear();
    }
}
int main(){
    while(~scanf("%d%d",&n,&q)){
        init();
        sz=round(sqrt(n));
        for(int i=1,u,v;i<n;i++){
            scanf("%d%d",&u,&v);
            edg[u].pb(v);edg[v].pb(u);
        }
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        dep[1]=1;
        dfs(1,0);
        while(q--){
            int u,v,k;
            scanf("%d%d%d",&u,&v,&k);
            int lca=LCA(u,v),ans=0;;
            if(k>sz){
                ans=solve(u,v,k,lca);
            }
            else {
                int dis=dep[u]-dep[lca];
                int s=(dis/k+1)*k;
                int x=find(u,s);
                ans=up[u][k]^up[x][k];
                int res=(dep[u]+dep[v]-2*dep[lca])%k;
                if(lca!=v&&dep[v]-dep[lca]>res){
                    v=find(v,res);
                    dis=dep[v]-dep[lca];
                    s=(dis/k+1)*k;
                    x=find(v,s);
                    ans^=up[v][k]^up[x][k];
                    if((dep[u]-dep[lca])%k==0&&(dep[v]-dep[lca])%k==0)ans^=a[lca];
                }
            }
            printf("%d
",ans);
        }
    }
    return 0;
}