LibreOJ #6284

题目链接:#6284. 数列分块入门 8

题目大意

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

solution

又是我们喜爱的分块

我们怎么做呢, 我们讲区间内全部相同的打一个标记, 记录这个区间的值,然后计算的时候加上这个区间的长度就好了,然后不相同的我们直接暴力查找就好了

我们在散块处理的时候记得把标记下放,不然会哭的

Code:

/**
*    Author: Alieme
*    Data: 2020.9.8
*    Problem: LibreOJ #6284
*    Time: O()
*/
#include <cstdio>
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>

#define int long long
#define rr register

#define inf 1e9
#define MAXN 200010

using namespace std;

inline int read() {
	int s = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= ch == '-', ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void print(int x) {
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) print(x / 10);
	putchar(x % 10 + 48);
}

int n, len;

int a[MAXN], id[MAXN], f[MAXN];

inline void reset(int x) {
	if (f[x] == -1) return ;
	for (rr int i = (x - 1) * len + 1; i <= min(x * len, n); i++) a[i] = f[x];
	f[x] = -1;
}

inline void change(int l, int r, int x) {
	int start = id[l], end = id[r];
	if (start == end) {
		reset(start);
		for (rr int i = l; i <= r; i++) a[i] = x;
		return ;
	}
	reset(start), reset(end);
	for (rr int i = l; id[i] == start; i++) a[i] = x;
	for (rr int i = start + 1; i < end; i++) f[i] = x;
	for (rr int i = r; id[i] == end; i--) a[i] = x;
}

inline int query(int l, int r, int x) {
	int start = id[l], end = id[r], ans = 0;
	if (start == end) {
		reset(start);
		for (rr int i = l; i <= r; i++) if (a[i] == x) ans++;
		return ans;
	}
	reset(start), reset(end);
	for (rr int i = l; id[i] == start; i++) if (a[i] == x) ans++;
	for (rr int i = start + 1; i < end; i++) 
		if (f[i] == x) ans += len;
		else if (f[i] == -1) for (rr int j = (i - 1) * len + 1; j <= i * len; j++) if (a[j] == x) ans++;
	for (rr int i = r; id[i] == end; i--) if (a[i] == x) ans++;
	return ans;
}

signed main() {
	// freopen("a1.in", "r", stdin);
	// freopen("a.out", "w", stdout);
	n = read();
	len = sqrt(n);
	for (rr int i = 1; i <= n; i++) a[i] = read(), id[i] = (i - 1) / len + 1, f[id[i]] = -1;
	for (rr int i = 1; i <= n; i++) {
		int l = read(), r = read(), c = read();
		cout << query(l, r, c) << "\n";
		change(l, r, c);
	}
}