Codeforces 221d D. Little Elephant and Array

最新推荐文章于 2020-07-04 16:14:23 发布

转载最新推荐文章于 2020-07-04 16:14:23 发布 · 120 阅读

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原文链接：http://www.cnblogs.com/ZlycerQan/p/6946285.html

本文介绍了一个基于 Codeforces221D 的题目“Little Elephant and Array”的解题思路，使用了莫队算法进行区间查询优化，并通过排序和区间移动来减少重复计算，最终实现了有效的解题方案。

二次联通门 : Codeforces 221d D. Little Elephant and Array

/*
    Codeforces 221d D. Little Elephant and Array
    
    题意 : 询问一段区间中出现次数等于自身的数的个数 
     
    正解是dp
    
    莫队水过， 作为我莫队的入门题 
    myj的思路 66
            
    把所有需查询的区间排序

    当前查询区间的答案为上一个区间的答案通过多次的区间移动得出

    
*/
#include <algorithm>
#include <cstdio>
#include <cmath>

#define Max 100005

void read (int &now)
{
    now = 0;
    register char word = getchar ();
    bool temp = false;
    while (word < '0' || word > '9')
    {
        if (word == '-')
            temp = true;
        word = getchar ();
    }
    while (word >= '0' && word <= '9')
    {
        now = now * 10 + word - '0';
        word = getchar ();
    }
    if (temp)
        now = -now;
}


int N, M;

int belong[Max];
struct Data
{
    int l, r;
    
    int ID;

    bool operator < (const Data &now) const 
    {
        if (belong[now.l] == belong[this->l])
            return now.r > this->r;
        return belong[now.l] > belong[this->l];
    }
};

int number[Max];

int K_Size;
int count[Max];
int rank_[Max];

Data query[Max];
int Answer[Max], Result;
int pos[Max];

incount void Updata (int now, int key)
{
    if (count[number[now]] == pos[now])
        Result--;
    count[number[now]] += key;
    if (count[number[now]] == pos[now])
        Result++;
}

int main (int argc, char *argv[])
{
    read (N);
    read (M);
    K_Size = sqrt (N);
    for (int i = 1; i <= N; i++)
    {
        read (number[i]);
        rank_[i] = number[i];
        pos[i] = number[i];
        belong[i] = (i + 1) / K_Size;
    }
    std :: sort (rank_ + 1, rank_ + 1 + N);
    int Size = std :: unique (rank_ + 1, rank_ + 1 + N) - rank_ - 1;
    for (int i = 1; i <= N; i++)
        number[i] = std :: lower_bound (rank_ + 1, rank_ + 1 + Size, number[i]) - rank_;
    for (int i = 1; i <= M; i++)
    {
        read (query[i].l);
        read (query[i].r);
        query[i].ID = i;    
    }
    std :: sort (query + 1, query + 1 + M);
    int l = 1, r = 0;
    for (int cur = 1, now_1, now_2; cur <= M; cur++)
    {
        now_1 = query[cur].l;
        now_2 = query[cur].r;
        if (l < now_1)
            for (int i = l; i < now_1; i++)
                Updata (i, -1);
        else
            for (int i = l - 1; i >= now_1; i--)
                Updata (i, 1);
        if (r < now_2)
            for (int i = r + 1; i <= now_2; i++)
                Updata (i, 1);
        else 
            for (int i = r; i > now_2; i--)
                Updata (i, -1);
        l = now_1;
        r = now_2;
        Answer[query[cur].ID] = Result;
    }
    for (int i = 1; i <= M; i++)
        printf ("%d\n", Answer[i]);
    return 0;
}

转载于:https://www.cnblogs.com/ZlycerQan/p/6946285.html