Article From:https://www.cnblogs.com/oi-forever/p/9125567.html

2120: Number of colors

Time Limit: 6 Sec  Memory Limit: 259 MB
Submit: 8079  Solved: 3321
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Description

Ink bought a set of N coloured paintbrushes (some of which may be the same color) and put them in a row. You need to answer ink and ink questions. Ink and ink will release the following instructions as you do: 1, Q L R representatives asked you to paint from the L brush to the R brush, there are several different colors of the brush. 2, R PCol replaced the P brush with color Col. Do you know what you need to do to meet ink requirements?

Input

The first row, two integers N and M, represent the number of initial brushes and the number of things that ink will do. The second line is N integers, representing the color of the I brush in the initial brush row. The third line to line 2+M, each row represents one thing that ink will do, and the format is dry.

Output

For each Query, you need to give a number in the corresponding line, representing the L brush to the R brush and there are several different colors in the brush.

Sample Input

6 5
1 2 3 4 5 5
Q 1 4
Q 2 6
R 1 2
Q 1 4
Q 2 6

Sample Output

4
4
3
4

HINT

 

For 100% of the data, N is less than 10000, M is less than 10000, and the modified operation is not more than 1000 times. All the integers in all the input data are equal to 1 and not more than 10^6.

Summary: to modify the Mo team board

#include <bits/stdc++.h>

using namespace std;
const int N = 1e5 + 7;
typedef long long ll;

struct Node{int l, r, tim, ID;} q[N];
struct DATA{int pos, Nw, Od;} c[N];
int t1, t2, sum[N * 10], blo, bl[N];
int n, m, l = 1, r, T, Ans, ans[N], s[N], now[N];
bool cmp(Node a, Node b) {
	return bl[a.l] == bl[b.l] ?(bl[a.r] == bl[b.r] ?a.tim < b.tim :a.r < b.r) :a.l < b.l;
}

void revise(int x, int d) {
	sum[x] += d;
	(d > 0) ?Ans += sum[x] == 1 :Ans -= sum[x] == 0;
}
void change(int x, int d) {
	if(l <= x && x <= r) {
		revise(d, 1); revise(s[x], -1);
	} s[x] = d;
}
int main() {
	char ch[2]; 
	scanf("%d%d", &n, &m); blo = sqrt(n);
	for (int i = 1; i <= n; ++i) scanf("%d", &s[i]), now[i] = s[i], bl[i] = i / blo  + 1;
	for (int i = 1; i <= m; ++i) {
		int x, y; scanf("%s%d%d", ch, &x, &y);
		if(ch[0] == 'Q') q[++t1] = (Node) {x, y, t2, t1};
		else c[++t2] = (DATA) {x, y, now[x]}, now[x] = y;
	}
	sort(q + 1, q + t1 + 1, cmp);
	for (int i = 1; i <= t1; ++i) {
		while(T < q[i].tim) change(c[T + 1].pos, c[T + 1].Nw), ++T;
		while(T > q[i].tim) change(c[T].pos, c[T].Od), --T;
		while(l < q[i].l) revise(s[l++], -1);
		while(l > q[i].l) revise(s[--l], 1);
		while(r < q[i].r) revise(s[++r], 1);
		while(r > q[i].r) revise(s[r--], -1);
		ans[q[i].ID] = Ans;
	}
	for (int i = 1; i <= t1; ++i) printf("%d\n", ans[i]);
	return 0;
}

  

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Link of this Article: BZOJ 2120: number color

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