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$O(n ^ 2)$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:***********************1599664856@qq.com**********************/#include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    #include 
    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 LOCALfreopen("/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;
    }

  
  

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