雅礼集训2018Day2 颜色

• 原题链接:雅礼集训 颜色

\(solution\)

• \(8 MB?\),还强制在线\(?\)若干个区间并\(?\)
• 看着极其不可做,没有任何数据结构能维护,考虑用\(bitset\)优化暴力(以下\(w = 64\)指\(bitset\)除掉的常数)
• 考虑分块\(bitset\),设块大小为\(S\),则块有\(\frac{n}{S}\)个,考虑朴素的时间复杂度单询问就是\(O(q * \frac{n}{w} * \frac{n}{s} + q * S)\),考虑做一些预处理
• 对于这种分块,我们的优化一般有两种手段线段树或者ST表,考虑这题是静态问题,且区间并不要求不交,考虑\(ST\)表
• 那么时间复杂度变为\(O(q * (\frac{n}{w} + S) + \frac{n^2}{Sw}*log(\frac{n}{S}))\),空间复杂度为\(O(\frac{n^2}{Sw}*log(\frac{n}{S}))\)
• 因为这题空间卡的很死,我们必须考虑不追求时间最优而用非常规大小分块,此处取\(S = \frac{n}{w}\),那么时间复杂度就是\(O(\frac{nq}{w} + n\log w)\),空间复杂度为\(O(n\log w)\)
• 代码先咕咕
• 本题卡常

/*雅礼集训2018 颜色*/ 
#include<bits/stdc++.h>
using namespace std;
int read(){
	char c = getchar();
	int x = 0;
	while(c < '0' || c > '9')	c = getchar();
	while(c >= '0' && c <= '9') x = x * 10 + c - 48,c = getchar();
	return x;
}
const int _ = 1e5 + 7;
int n,m,p;
int a[_];int L[69],R[69]; 
int b[_];
bitset<_>o[69][7],Ans;int lg2[69];
int S,t;
void solve(int l,int r){
//	cout<<b[l]<<' '<<b[r]<<endl;
	if(b[l] == b[r]){
		for(int i = l; i <= r; ++i)		Ans.set(a[i]);
		return;
	}
//	cout<<l<<' '<<r<<' '<<b[l]<<' '<<b[r]<<"!!"<<endl;
	for(int i = l; i <= R[b[l]]; ++i)	Ans.set(a[i]);
	for(int i = L[b[r]]; i <= r; ++i)	Ans.set(a[i]);
	l = b[l] + 1,r = b[r] - 1;
	if(l <= r){
		int k = lg2[r-l+1];
//		assert(k <= 6);
//		assert(k >= 0);
		Ans |= o[l][k];Ans |= o[r-(1<<k)+1][k];
	}
}
int main(){
//	freopen("color.in","r",stdin);
//	freopen("color.out","w",stdout);
	n = read(),m = read(),p = read();
	for(int i = 1; i <= n; ++i)		a[i] = read();
	for(int i = 2; i <= 65; ++i)	lg2[i] = lg2[i>>1] + 1;
	int ans = 0;
	if(n < 64){
		for(int T = 1; T <= m; ++T){
			int k = read();
			Ans.reset();
			while(k--){
				int l = read(),r = read();
				if(p && T != 1){
					l = (l ^ ans) % n + 1;
					r = (r ^ ans) % n + 1;
				}
//				cout<<l<<' '<<r<<endl;
				for(int i = l; i <= r; ++i)	Ans.set(a[i]);
			}
			ans = Ans.count();
			printf("%d\n",ans);
		}		
		return 0;
	}
	S = n / 64;t = 0;
	for(int i = 1; i <= 64; ++i){
		++t;
		L[t] = R[t-1] + 1;
		R[t] = L[t] + S - 1;
	}
	if(R[t] != n){
		++t;L[t] = R[t-1] + 1;
		R[t] = n;
	}
//	cout<<t<<' '<<S<<endl;
	for(int i = 1; i <= t; ++i){
		for(int j = L[i]; j <= R[i]; ++j){
			o[i][0].set(a[j]),b[j] = i;
		}
	}
	for(int j = 1; j <= 6; ++j)
		for(int i = 1; i + (1 << j) - 1 <= t; ++i)
			o[i][j] = o[i][j-1] | o[i+(1<<(j-1))][j-1];
//	int ans = 0;
	for(int T = 1; T <= m; ++T){
		int k = read();
		Ans.reset();
		while(k--){
			int l = read(),r = read();
			if(p && T != 1){
				l = (l ^ ans) % n + 1;
				r = (r ^ ans) % n + 1;
			}
			if(l > r)	swap(l,r);
			solve(l,r);
		}	
		ans = Ans.count();
		printf("%d\n",ans);
	}
	return 0;
}