Solution - P1903 【模板】带修莫队 / [国家集训队] 数颜色 / 维护队列

带修莫队。那是什么。能吃吗。

我们可以把区间数颜色转化为二维数点，带修就是动态二位数点，即三维数点。

然后随便做了。

#include <bits/stdc++.h>
#define llong long long
#define N 4000006
using namespace std;

#define bs (1<<20)
char buf[bs], *p1, *p2;
#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,bs,stdin),p1==p2)?EOF:*p1++)
template<typename T>
inline void read(T& x){
	x = 0; int w = 1;
	char ch = gc();
	while(ch < '0' || ch > '9'){
		if(ch == '-') w = -w;
		ch = gc();
	}
	while(ch >= '0' && ch <= '9')
		x = (x<<3)+(x<<1)+(ch^48), ch = gc();
	x *= w;
}
inline void read(char& ch){
	ch = gc();
	while(ch == ' ' || ch == '\r' || ch == '\n') ch = gc();
}
template<typename T, typename ...Args>
inline void read(T& x, Args& ...y){
	return read(x), read(y...);
}

int n, q, cnt, qcnt;
int b[N];
struct Node{
	int t, x, y, w;
	int id;
} a[N];
int ans[N];

set<int> col[N];

int fwk[N];
#define lowbit(x) (x&-x)
inline void modify(int x, int k){for(int i = x; i < N; i += lowbit(i)) fwk[i] += k;}
inline int query(int x){int res = 0; for(int i = x; i; i ^= lowbit(i)) res += fwk[i]; return res;}
inline void clear(int x){for(int i = x; i < N; i += lowbit(i)) fwk[i] = 0;}

#define mid ((l+r)>>1)
inline void cdq(int l, int r){
	if(l == r) return;
	cdq(l, mid), cdq(mid+1, r);
	sort(a+l, a+mid+1,   [&](Node o1, Node o2){return o1.x<o2.x;});
	sort(a+mid+1, a+r+1, [&](Node o1, Node o2){return o1.x<o2.x;});
	int i = l, j = mid+1;
	while(j <= r){
		while(i <= mid && a[i].x <= a[j].x){
			if(!a[i].id) modify(a[i].y, a[i].w);
			++i;
		}
		if(a[j].id) ans[a[j].id] += query(a[j].y)*a[j].w;
		++j;
	}
	for(int k = l; k < i; ++k)
		if(!a[k].id) clear(a[k].y);
	return;
}

int main(){
	freopen("in.txt", "r", stdin);
	read(n, q);
	for(int i = 1; i <= n; ++i) read(b[i]);
	for(int i = n; i >= 1; --i){
		a[++cnt] = {0, i, i, 1, 0};
		auto it = col[b[i]].upper_bound(i);
		if(it != col[b[i]].end()){
			int pos = *it;
			a[++cnt] = {0, pos, i, -1, 0};
		}
		col[b[i]].insert(i);
	}
	for(int i = 1; i <= q; ++i){
		char op; int x, y;
		read(op, x, y);
		if(op == 'R'){
			// Erase
			int pos1 = 0, pos2 = 0;
			set<int>::iterator it1, it2;
			it1 = it2 = col[b[x]].lower_bound(x);
			if(it1 != col[b[x]].begin()) pos1 = *--it1;
			if(++it2 != col[b[x]].end()) pos2 = *it2;
			if(pos1) a[++cnt] = {i, pos1, x, 1, 0};
			if(pos2) a[++cnt] = {i, x, pos2, 1, 0};
			if(pos1 && pos2) a[++cnt] = {i, pos1, pos2, -1, 0};
			col[b[x]].erase(x);
			// Insert
			b[x] = y, pos1 = pos2 = 0;
			it1 = it2 = col[b[x]].lower_bound(x);
			if(it1 != col[b[x]].begin()) pos1 = *--it1;
			if(it2 != col[b[x]].end())   pos2 = *it2;
			if(pos1) a[++cnt] = {i, pos1, x, -1, 0};
			if(pos2) a[++cnt] = {i, x, pos2, -1, 0};
			if(pos1 && pos2) a[++cnt] = {i, pos1, pos2, 1, 0};
			col[b[x]].insert(x);
		}
		if(op == 'Q'){
			++qcnt;
			a[++cnt] = {i, y,   y,   1,  qcnt};
			if(x == 1) continue;
			a[++cnt] = {i, x-1, y,   -1, qcnt};
			a[++cnt] = {i, y,   x-1, -1, qcnt};
			a[++cnt] = {i, x-1, x-1, 1,  qcnt};
		}
	}
	sort(a+1, a+cnt+1, [&](Node o1, Node o2){return o1.t<o2.t || (o1.t==o2.t&&(o1.x<o2.x||(o1.x==o2.x&&(o1.y<o2.y||(o1.y==o2.y&&o1.id<o2.id)))));});
	cdq(1, cnt);
	for(int i = 1; i <= qcnt; ++i) printf("%d\n", ans[i]);
	return 0;
}