LOJ6284 数列分块入门8

LOJ6284 数列分块入门 8

标签

• 分块入门

前言

• 我的csdn和博客园是同步的，欢迎来访danzh-博客园~

简明题意

• 维护序列，支持两种操作：
  1. 查询区间中某种权值的出现次数
  2. 将整个区间[l,r]全部改为c

思路

• 这个暴力得想不到呀。
• 真的很暴力，实际上开一个数组去维护一下每一块的值是否是一样的，再开一个数组记录如果某一块的值相同那么这个值是多少。然后就很简单了惹

注意事项

• 无

总结

• 无

AC代码

#pragma GCC optimize(2)
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;

const int maxn = 1e5 + 10;

int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
	while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}

int n, a[maxn];
int pos[maxn], len, val[maxn];
bool is_zh[maxn];

int cal(int k, int c)
{
	if (is_zh[k])
		return val[k] == c ? len : 0;

	int ans = 0;
	for (int i = k * len - len + 1; i <= k * len; i++)
		if (a[i] == c)
			ans++;
	return ans;
}

void reset(int k)
{
	if (is_zh[k])
	{
		for (int i = k * len - len + 1; i <= min(k * len, n); i++)
			a[i] = val[k];
		is_zh[k] = 0;
	}
}

int deal(int l, int r, int c)
{
	//查询
	int ans = 0;
	reset(pos[l]);
	for (int i = l; i <= min(pos[l] * len, r); i++)
		if (a[i] == c)
			ans++;

	if (pos[l] != pos[r])
	{
		reset(pos[r]);
		for (int i = pos[r] * len - len + 1; i <= r; i++)
			if (a[i] == c)
				ans++;
	}
	for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
		ans += cal(i, c);
	
	//修改
	
	for (int i = l; i <= min(pos[l] * len, r); i++)
		a[i] = c;
	if (pos[l] != pos[r])
	{
		
		for (int i = pos[r] * len - len + 1; i <= r; i++)
			a[i] = c;
	}
	for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
		is_zh[i] = 1, val[i] = c;

	return ans;
}

void solve() {
	scanf("%d", &n);
	len = sqrt(n);
	for (int i = 1; i <= n; i++)
	a[i] = read(), pos[i] = (i - 1) / len + 1;

	for (int i = 1; i <= n; i++) {
		int l, r, c;
		l = read(), r = read(), c = read();

		printf("%d\n", deal(l, r, c));
	}
}

int main() {
	freopen("Testin.txt", "r", stdin);
	freopen("Testout.txt", "w", stdout);
	solve();
	return 0;
}