区间众数

1. luogu P5048

大意: 区间询问众数的出现次数, 强制在线.

先预处理出块间答案, 考虑每次询问左右边界, 显然最多使答案再增加$2sqrt{n}$. 预处理时把相同元素按顺序存进vector里, 这样可以O(1)查询每个元素前后k次出现的位置, 对于左边界若后ans次出现位置<=r, 则++ans, 右边界若前ans次出现位置>=l, 则++ans

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 5e5+10, M = 1000;
int n, m, cnt, sz, ans;
map<int,int> s;
vector<int> val[N];
int a[N], sum[N], blo[N], pos[N];
int L[M], R[M], f[M][M];
int id(int x) {
	if (s.count(x)) return s[x];
	int t = s.size();
	return s[x]=t;
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) {
		scanf("%d", a+i);
		a[i] = id(a[i]);
		val[a[i]].pb(i);
		pos[i] = val[a[i]].size()-1;
	}
	sz = 710, cnt = (n-1)/sz+1;
	REP(i,1,cnt) L[i]=R[i-1]+1,R[i]=i*sz;
	R[cnt] = n;
	REP(i,1,cnt) {
		memset(sum,0,sizeof sum);
		REP(j,L[i],R[i]) blo[j]=i;
		REP(j,i,cnt) {
			int &res = f[i][j] = f[i][j-1];
			REP(k,L[j],R[j]) {
				++sum[a[k]];
				res = max(res, sum[a[k]]);
			}
		}
	}
	memset(sum,0,sizeof sum);
	REP(i,1,m) {
		int l, r;
		scanf("%d%d", &l, &r);
		l ^= ans, r ^= ans, ans = f[blo[l]+1][blo[r]-1];
		if (blo[l]==blo[r]) {
			REP(i,l,r) {
				++sum[a[i]];
				ans=max(ans,sum[a[i]]);
			}
			REP(i,l,r) sum[a[i]]=0;
		}
		else {
			REP(i,l,R[blo[l]]) {
				int res = pos[i], sz = val[a[i]].size();
				while (res+ans<sz&&val[a[i]][res+ans]<=r) ++ans;
			}
			REP(i,L[blo[r]],r) {
				int res = pos[i];
				while (res-ans>=0&&val[a[i]][res-ans]>=l) ++ans;
			}
		}
		printf("%d
", ans);
	}
}

2. luogu P4168

大意: 区间询问最小众数, 强制在线.

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '
'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 5e5+10, M = 1000;
int n, m, cnt, sz, ans;
map<int,int> s;
vector<int> val[N];
int a[N], b[N], sum[N], blo[N], pos[N];
int L[M], R[M], f[M][M], g[M][M];
int id(int x) {
	if (s.count(x)) return s[x];
	int t = s.size()+1;
	b[t] = x;
	return s[x]=t;
}

int main() {
	scanf("%d%d", &n, &m);
	REP(i,1,n) {
		scanf("%d", a+i);
		a[i] = id(a[i]);
		val[a[i]].pb(i);
		pos[i] = val[a[i]].size()-1;
	}
	sz = 700, cnt = (n-1)/sz+1;
	REP(i,1,cnt) L[i]=R[i-1]+1,R[i]=i*sz;
	R[cnt] = n;
	REP(i,1,cnt) {
		memset(sum,0,sizeof sum);
		REP(j,L[i],R[i]) blo[j]=i;
		int ans = 0, mx = 0;
		REP(j,i,cnt) {
			REP(k,L[j],R[j]) {
				++sum[a[k]];
				if (sum[a[k]]>mx||sum[a[k]]==mx&&b[a[k]]<b[a[ans]]) ans=k,mx=sum[a[k]];
			}
			f[i][j] = ans, g[i][j] = mx;
		}
	}
	memset(sum,0,sizeof sum);
	REP(i,1,m) {
		int l, r;
		scanf("%d%d", &l, &r);
		l = (l+ans-1)%n+1, r = (r+ans-1)%n+1;
		if (l>r) swap(l,r);
		ans = f[blo[l]+1][blo[r]-1];
		int mx = g[blo[l]+1][blo[r]-1];
		if (blo[l]==blo[r]) {
			REP(i,l,r) {
				++sum[a[i]];
				if (sum[a[i]]>mx||sum[a[i]]==mx&&b[a[i]]<b[a[ans]]) ans=i,mx=sum[a[i]];
			}
			REP(i,l,r) sum[a[i]]=0;
		}
		else {
			REP(i,l,R[blo[l]]) {
				int res = pos[i], sz = val[a[i]].size(), p = res+mx;
				if (0<=p-1&&p-1<sz&&val[a[i]][p-1]<=r) {
					if (b[a[i]]<b[a[ans]]) ans=i;	
				}
				while (p<sz&&val[a[i]][p]<=r) ++p, ++mx, ans = i;
			}
			REP(i,L[blo[r]],r) {
				int res = pos[i], sz = val[a[i]].size(), p = res-mx;
				if (0<=p+1&&p+1<sz&&val[a[i]][p+1]>=l) {
					if (b[a[i]]<b[a[ans]]) ans=i;
				}
				while (p>=0&&val[a[i]][p]>=l) --p, ++mx, ans = i;
			}
		}
		printf("%d
", ans=b[a[ans]]);
	}
}