LOJ-6282-数列分块入门6

链接:

https://loj.ac/problem/6282

题意:

给出一个长为 的数列，以及 个操作，操作涉及单点插入，单点询问，数据随机生成。

思路:

vector 维护每个区间, 当某个区间的值太多时,重构一下.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int MAXN = 1e5+10;

int Belong[MAXN*10], a[MAXN*10];
vector<int> Vec[MAXN];
int n, part, last;

pair<int, int> GetPos(int p)
{
    int p1 = 1;
    while (p > Vec[p1].size())
        p -= Vec[p1].size(), p1++;
    return make_pair(p1, p-1);
}

void Rebuild()
{
    int pos = 0;
    for (int i = 1;i <= last;i++)
    {
        for (vector<int>::iterator it = Vec[i].begin();it != Vec[i].end();++it)
            a[++pos] = *it;
        Vec[i].clear();
    }
    part = sqrt(pos);
    last = (pos-1)/part+1;
    for (int i = 1;i <= pos;i++)
    {
        Belong[i] = (i-1)/part+1;
        Vec[Belong[i]].push_back(a[i]);
    }
}

void Update(int l, int r, int c)
{
    pair<int, int> p = GetPos(l);
    Vec[p.first].insert(Vec[p.first].begin()+p.second, r);
    if (Vec[p.first].size() > 10*part)
        Rebuild();
}

int Query(int l, int r, int c)
{
    pair<int, int> p = GetPos(r);
    return Vec[p.first][p.second];
}

int main()
{
    scanf("%d", &n);
    part = sqrt(n);
    int v;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d", &v);
        Belong[i] = (i-1)/part+1;
        Vec[Belong[i]].push_back(v);
    }
    last = (n-1)/part+1;//最后一个区间
    int op, l, r, c;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d", &op);
        if (op == 0)
        {
            scanf("%d%d%d", &l, &r, &c);
            Update(l, r, c);
        }
        else
        {
            scanf("%d%d%d", &l, &r, &c);
            printf("%d\n", Query(l, r, c));
        }
    }

    return 0;
}