Xor Sum （01字典树）基础的字典树题目

初学字典树的可以拿这道题练练手。

题目链接：传送门

题目大意：给你n个数，有m次询问，每次给个数，问：这个数与n个数中的哪个数的异或值最大，输出它。

给大家两种写法；
1。数组的写法：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;//typedef __int64 ll;typedef long long ll;const int M = 55;const int N = M*1e5;struct Node{    ll val;    int l;    int r;    void clear()    {        l = r = -1;    }} node[N];int p;ll a, t[M];void insert (int& root, int d, ll u){    if (root == -1)    {        root = p++;        node[root].clear();    }    if (d == -1)    {        node[root].val = u;        return;    }    if (u & t[d])        insert(node[root].r, d - 1, u);    else        insert(node[root].l, d - 1, u);}void query(int root, int d, ll u){    if (d == -1)    {        printf("%lld\n", node[root].val);        return;    }    if (((u & t[d]) && node[root].l != -1) || node[root].r == -1)        query(node[root].l, d - 1, u);    else        query(node[root].r, d - 1, u);}int main (){    int cas, n, m;    scanf("%d", &cas);    t[0] = 1;    for (int i = 1; i < 55; i++)        t[i] = t[i-1] * 2;    for (int i = 1; i <= cas; i++)    {        p = 0;        int root = -1;        scanf("%d%d", &n, &m);        for (int j = 0; j < n; j++)        {            scanf("%lld", &a);            insert(root, 50, a);        }        printf("Case #%d:\n", i);        for (int j = 0; j < m; j++)        {            scanf("%lld", &a);            query(root, 50, a);        }    }    return 0;}

2。指针的写法：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;ll a,wei[56];struct node{    node *left,*right;    bool l,r;    ll data;    void clean()    {        l=0;        data=0;        r=0;    }};void build(node *head,int f){    if(f==-1)    {        head->data=a;        return ;    }    if(a&wei[f])    {        if(head->r==0)        {            head->r=1;            head->right=new node;            head->right->clean();        }        build(head->right,f-1);    }    else    {        if(head->l==0)        {        head->l=1;            head->left=new node;            head->left->clean();        }        build(head->left,f-1);    }}void query(node *head,ll f){    if(f==-1)    {        printf("%lld\n",head->data);        return ;    }    //printf("                   fjhd\n");    if(((a&wei[f])&&head->l==1)||head->r==0)        query(head->left,f-1);    else query(head->right,f-1);}int main(){    int t,tt=1;    wei[0]=1;    for(int i=1; i<=55; i++)        wei[i]=wei[i-1]*2;    scanf("%d",&t);    while(t--)    {        int n,m;        scanf("%d%d",&n,&m);        node *head=new node;        head->clean();        for(int i=0; i<n; i++)        {            scanf("%lld",&a);            build(head,50);        }        printf("Case #%d:\n",tt++);        for(int i=0; i<m; i++)        {            scanf("%lld",&a);            query(head,50);        }    }    return 0;}