SPOJ - SUBXOR（01字典树）

来源：互联网发布：淘宝分销商发货流程编辑：程序博客网时间：2024/05/31 05:27

SUBXOR - SubXor

A straightforward question. Given an array of positive integers you have to print the number of subarrays whose XOR is less than K.
Subarrays are defined as a sequence of continuous elements Ai, Ai+1, ..., Aj . XOR of a subarray is defined as Ai^Ai+1^ ... ^Aj.
Symbol ^ is Exclusive Or. You can read more about it here:
http://en.wikipedia.org/wiki/Exclusive_or

Input Format:

First line contains T, the number of test cases. Each of the test case consists of N and K in one line, followed by N space separated integers in next line.

Output Format:

For each test case, print the required answer.

Constraints:

1 ≤ T ≤ 10
1 ≤ N ≤ 10^5
1 ≤ A[i] ≤ 10^5
1 ≤ K ≤ 10^6
Sum of N over all testcases will not exceed 10^5.

Sample Input:

15 24 1 3 2 7

Sample Output:

问题概述：给你一个长度为n的整数序列和一个k，问这个序列中有多少子串的异或和<k（一个子串ai,ai+1……aj的

异或和指的是ai^ai+1^……^aj）

https://vjudge.net/contest/145297#problem/D

异或性质：

①：a^b==c，那么a^c==b

②：a^b^b==a

解题步骤：

①：求出所有异或前缀和b[i]

②：枚举每一个b[i]，对于当前b[i]，将b[i-1]加入01字典树（注意如果当前是b[1]，将"b[0]==0"加入字典树），之后

查询字典树找到满足b[j]^b[i]<k(0<j<i)的有多少个，最后将结果加在一起即可

字典树怎么建？

每个节点存储一个sum值，表示b[]中对应前缀数量（即有多少个b[j]在加入字典树时经过了这个节点）

怎么查询？

①：尽可能找到一条路径x，使得x^b[i]==k，到某个节点无法往下走了就直接返回ans

②：寻找的过程中如果k的对应位是1那么ans += sum[p]，其中p^b[i]==0，p为节点上的数值(0or1)<<(最高位-树深)

#include<stdio.h>#include<string.h>typedef struct Trie{int sum;Trie *next[2];}Trie;int k, a[100005], b[100005], vc[25];void Create(Trie *root, int a){int now, i;Trie *p = root;for(i=24;i>=0;i--){now = 0;if(a&(1<<i))now = 1;if(p->next[now]==NULL){Trie *temp = new Trie;temp->next[0] = temp->next[1] = NULL;temp->sum = 0;p->next[now] = temp;}p = p->next[now];p->sum++;}}int Query(Trie *root, int a){int now, i, ans;ans = 0;Trie *p = root;for(i=24;i>=0;i--){now = 0;if(a&(1<<i))now = 1;if(vc[i]==1){if(p->next[!(1^now)]!=NULL)ans += p->next[!(1^now)]->sum;}if(p->next[vc[i]^now]==NULL)return ans;p = p->next[vc[i]^now];}return ans;}void Delete(Trie *root){if(root->next[0])Delete(root->next[0]);if(root->next[1])Delete(root->next[1]);delete root;}int main(void){int T, i, n, k;long long ans;scanf("%d", &T);while(T--){Trie *root = new Trie;root->next[0] = root->next[1] = NULL;root->sum = 0;ans = 0;scanf("%d%d", &n, &k);for(i=1;i<=n;i++){scanf("%d", &a[i]);b[i] = b[i-1]^a[i];}memset(vc, 0, sizeof(vc));for(i=24;i>=0;i--){if(k&(1<<i))vc[i] = 1;}for(i=1;i<=n;i++){Create(root, b[i-1]);ans += Query(root, b[i]);}printf("%lld\n", ans);Delete(root);}return 0;}

1 0