cf665e.cpp 01 trie树求抑或

题意:给定一个数列,求其中抑或值大于等于k的子序列的数量.
分析:

• 首先知道抑或有类似于前缀和的区间性质,可以O(n2)求任意区间抑或值.
• 预处理前缀抑或,然后对以每个数结尾的抑或都去查其前面的子序列,显然这里复杂度是平方.
• 然后我们就想到用trie树来维护一颗01字典树的经典用法,每次去查询当前点之前的前缀即可,而且可以有两个
  剪枝
• • 最优化剪枝,如果之后的抑或值全为1都不能大于等于k就减去
• • 如果当前前缀已经满足,就直接加前缀值,就不用查后面.

/**********************jibancanyang**************************  * *Author        :jibancanyang  *Created Time  : 五  4/22 11:23:34 2016 *File Name     : cf665e.cpp *Problem:01 trie树求抑或 *Get: ***********************[e-mail]**********************/#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;vector<int> vi;#define pr(x) cout << #x << ": " << x << "  "#define pl(x) cout << #x << ": " << x << endl;#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define rep(i, a, n) for (int i = a; i < n; i++)#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++)const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e6 + 12;int xors[maxn], k, n;int maxb = 28;struct node {    node *l, *r;    ll cnt;    node(void): l(NULL), r(NULL), cnt(0) {}} *root;inline int setone(int x, int temp) {    return x | (1 << temp); }inline int setzero(int x, int temp) {    return x & (~(1 << temp));}void add(int x, node *cur) {    for (int i = maxb; i >= 0; i--) {        bool temp = (x >> i) & 1;        if (temp) {            if (cur -> r == NULL) cur -> r = new node;            cur = cur -> r;        } else {            if (cur -> l == NULL) cur -> l = new node;            cur = cur -> l;        }        (cur -> cnt)++;    }}ll search(int x, node *cur, int t, int op) {    if ((t ^ x) + (1 << (op + 1)) < k) return 0;    ll ret = 0;    if (cur != root && (t ^ x) >= k) {        ret += cur -> cnt;        return ret;    }    if (cur -> l != NULL) ret += search(x, cur -> l, setzero(t, op), op - 1);    if (cur -> r != NULL) ret += search(x, cur -> r, setone(t, op), op - 1);    return ret;}int main(void){#ifdef LOCAL    freopen("/Users/zhaoyang/in.txt", "r", stdin);    //freopen("/Users/zhaoyang/out.txt", "w", stdout);#endif    //ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);    cin >> n >> k;    xors[0] = 0;    root = new node;    rep (i, 1, n + 1) sa(xors[i]),  xors[i] = xors[i] ^ xors[i - 1];    ll ans = 0;    rep (i, 1, n + 1) {        // pl(i);       add(xors[i - 1], root);       ans += search(xors[i], root, xors[i], maxb);    }    cout << ans << endl;    return 0;}