LOJ-6284-数列分块入门8

链接:

https://loj.ac/problem/6284

题意:

给出一个长为 的数列，以及 个操作，操作涉及区间询问等于一个数 的元素，并将这个区间的所有元素改为 。

思路:

维护一个分块是否全部等于一个值,不等于时跑暴力即可.

查询一段时先把对应的分块更新再查询和更改.

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;
const int MOD = 10007;

int a[MAXN], Tag[MAXN];
int Belong[MAXN];
int n, part, last;

void Reset(int pos)
{
    if (Tag[pos] == -1)
        return;
    for (int i = (pos-1)*part+1;i <= pos*part;i++)
        a[i] = Tag[pos];
    Tag[pos] = -1;
}

int Solve(int l, int r, int c)
{
    int cnt = 0;
    Reset(Belong[l]);
    for (int i = l;i <= min(Belong[l]*part, r);i++)
    {
        if (a[i] == c)
            cnt++;
        a[i] = c;
    }
    if (Belong[l] != Belong[r])
    {
        Reset(Belong[r]);
        for (int i = max((Belong[r]-1)*part+1, l);i <= r;i++)
        {
            if (a[i] == c)
                cnt++;
            a[i] = c;
        }
    }
    for (int i = Belong[l]+1;i <= Belong[r]-1;i++)
    {
        if (Tag[i] != -1)
        {
            if (Tag[i] == c)
                cnt += part;
            Tag[i] = c;
        }
        else
        {
            for (int j = (i-1)*part+1;j <= i*part;j++)
            {
                if (a[j] == c)
                    cnt++;
                a[j] = c;
            }
            Tag[i] = c;
        }
    }
    return cnt;
}

int main()
{
    scanf("%d", &n);
    part = sqrt(n);
    memset(Tag, -1, sizeof(Tag));
    for (int i = 1;i <= n;i++)
    {
        scanf("%d", &a[i]);
        Belong[i] = (i-1)/part+1;
    }
    int op, l, r, c;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d%d%d", &l, &r, &c);
        printf("%d\n", Solve(l, r, c));
    }

    return 0;
}