[TJOI 2018] XOR

[题目链接]

       https://www.lydsy.com/JudgeOnline/problem.php?id=5338

[算法]

        首先对这棵树进行树链剖分

        那么我们就将一个树上的问题转化为一个序列上的问题

        建立可持久化字典树维护最大异或值即可

        时间复杂度 : O(NlogN ^ 2)

[代码]

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int N = 2e5 + 10;
const int MAXLOG = 31;

struct edge
{
        int to , nxt;
} e[N << 1];

int n , tot , timer , q;
int head[N] , size[N] , top[N] , a[N] , perm[N] , l[N] , r[N] , son[N] , father[N] , depth[N] , rt[N];

struct Presitent_Trie
{
        int sz;
        int child[N * MAXLOG][2] , latest[N * MAXLOG];
        Presitent_Trie()
        {
                sz = 0;        
        }        
        inline void modify(int bit , int &now , int old , int x , int loc)
        {
                now = ++sz;
                child[now][0] = child[old][0] , child[now][1] = child[old][1];
                latest[now] = loc;
                if (bit < 0) return;
                int value = 0;
                if (x & (1 << bit)) value = 1;
                modify(bit - 1 , child[now][value] , child[old][value] , x , loc);
        }
        inline int query(int bit , int now , int lft , int x)
        {
                if (bit < 0)
                        return 0;
                int value = 1;
                if (x & (1 << bit)) value = 0;
                if (latest[child[now][value]] >= lft) return (1 << bit) + query(bit - 1 , child[now][value] , lft , x);
                else return query(bit - 1 , child[now][value ^ 1] , lft , x);
        }
} PT;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
inline void addedge(int u , int v)
{
        ++tot;
        e[tot] = (edge){v , head[u]};
        head[u] = tot;
}
inline void dfs1(int u , int par)
{
        size[u] = 1;
        depth[u] = depth[par] + 1;
        father[u] = par;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (v == par) continue;
                dfs1(v , u);
                size[u] += size[v];
                if (size[v] > size[son[u]]) son[u] = v;
        }        
}
inline void dfs2(int u , int t)
{
        top[u] = t;
        l[u] = ++timer;
        if (son[u]) dfs2(son[u] , t);
        for (int i = head[u]; i; i = e[i].nxt)        
        {
                int v = e[i].to;
                if (v == father[u] || v == son[u]) continue;
                dfs2(v , v);
        }
        r[u] = timer;
}
inline bool cmp(int x , int y)
{
        return l[x] < l[y];
} 
inline int query(int x , int y , int z)
{    
        int ans = 0;
        while (top[x] != top[y])
        {
                if (depth[top[x]] > depth[top[y]]) swap(x , y);
                chkmax(ans , PT.query(MAXLOG - 1 , rt[l[y]] , l[top[y]] , z));
                y = father[top[y]];
        }
        if (depth[x] > depth[y]) swap(x , y);
        chkmax(ans , PT.query(MAXLOG - 1 , rt[l[y]] , l[x] , z));
        return ans;
}

int main()
{
        
        read(n); read(q);
        for (int i = 1; i <= n; ++i) read(a[i]);
        for (int i = 1; i < n; ++i)
        {
                int x , y;
                read(x); read(y);
                addedge(x , y);
                addedge(y , x);
        }
        dfs1(1 , 0);
        dfs2(1 , 1);
        for (int i = 1; i <= n; ++i) perm[i] = i;
        sort(perm + 1 , perm + n + 1 , cmp);
        for (int i = 1; i <= n; ++i) PT.modify(MAXLOG - 1 , rt[i] , rt[i - 1] , a[perm[i]] , i); 
        while (q--) 
        {
                int type;
                read(type);
                if (type == 1)
                {
                        int x , y;
                        read(x); read(y);
                        printf("%d\n" , PT.query(MAXLOG - 1 , rt[r[x]] , l[x] , y));        
                }    else
                {
                        int x , y , z;
                        read(x); read(y); read(z);
                        printf("%d\n" , query(x , y , z));
                }
        }
        
        return 0;
    
}

标签：XOR,int,TJOI,long,son,void,2018,inline,top
来源： https://www.cnblogs.com/evenbao/p/10540007.html