• BZOJ 1878 HH的项链


    莫队

    还是一道模板。。不过洛谷数据加强了,必须要奇偶性排序+吸氧才能过,BZOJ可以直接过的~

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int X = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
        return w ? -X : X;
    }
    inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template<typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template<typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template<typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 500005;
    const int M = 1000005;
    
    int n, m, t, a[N], freq[M], ans, res[N];
    struct Query{
        int l, r, id, block;
        bool operator < (const Query &rhs) const {
            return (block ^ rhs.block) ? l < rhs.l : (block & 1) ? r < rhs.r : r > rhs.r;
        }
    }query[N];
    
    inline void add(int k){
        freq[a[k]] ++;
        if(freq[a[k]] == 1) ans ++;
    }
    
    inline void remove(int k){
        freq[a[k]] --;
        if(freq[a[k]] == 0) ans --;
    }
    
    int main(){
    
        n = read();
        for(int i = 1; i <= n; i ++) a[i] = read();
        m = read();
        t = (int)sqrt(n);
        for(int i = 1; i <= m; i ++){
            query[i].l = read(), query[i].r = read();
            query[i].id = i, query[i].block = (query[i].l - 1) / t + 1;
        }
        sort(query + 1, query + m + 1);
        int l = 1, r = 0;
        for(int i = 1; i <= m; i ++){
            int curL = query[i].l, curR = query[i].r;
            while(l < curL) remove(l ++);
            while(l > curL) add(-- l);
            while(r < curR) add(++ r);
            while(r > curR) remove(r --);
            res[query[i].id] = ans;
        }
        for(int i = 1; i <= m; i ++){
            printf("%d
    ", res[i]);
        }
        return 0;
    }
    
  • 相关阅读:
    SQL分类
    广度/深度优先生成树
    图的基本概念
    哈夫曼树构造/哈夫曼编码
    二叉排序树/平衡二叉树
    树、森林与二叉树的转换
    树/二叉树的基本性质
    /*传说中的土办法找中序前驱*/
    KiCAD原理图更换库
    博客园添加版权信息
  • 原文地址:https://www.cnblogs.com/onionQAQ/p/10858389.html
Copyright © 2020-2023  润新知