BZOJ 3339 Rmq Problem【離線，值域線段樹

區間mex，考慮離線按左端點排序

左端點右移時，這個數到它的下一個數之間的位置（作爲右端點）的mex值全部與當前數取min

#include<bits/stdc++.h>
#define MAXN 200005
#define MX 200005
#define INF 0x3f3f3f3f
using namespace std;	int n,m;
inline int read(){
	char ch=getchar();
	while(!isdigit(ch))	ch=getchar();
	int rtn = 0;
	while(isdigit(ch))	rtn = rtn*10 +ch-'0' , ch=getchar();
	return rtn;
}
int a[MAXN];
int mex[MAXN],tag[MAXN<<2];
int nxt[MAXN],lst[MAXN];

struct t1{
	int l,r,id;
	bool operator < (const t1 &ano) const{
		return l<ano.l;
	}
}Q[MAXN];
int ans[MAXN];

void modify(int now,int l,int r,int L,int R,int v){
	if(L<=l&&r<=R)	return void(tag[now] = min(tag[now],v));
	int mid = (l+r)>>1;
	if(L<=mid)	modify(now<<1,l,mid,L,R,v);
	if(mid<R)	modify(now<<1|1,mid+1,r,L,R,v);
}

int inqry(int now,int l,int r,int pos){
	int rtn = tag[now];
	if(l==r)	return rtn = min(rtn,mex[l]);
	int mid = (l+r)>>1;
	if(pos<=mid)	return min(rtn,inqry(now<<1,l,mid,pos));
	else	return min(rtn,inqry(now<<1|1,mid+1,r,pos));
}

int main(){
//	freopen("1.in","r",stdin);
	
	n=read(),m=read();
	for(int i=1;i<=n;++i)	a[i]=read();

	for(int i=1,k=0;i<=n;++i){
		tag[a[i]] = 1;
		while(tag[k])	++k;
		mex[i] = k;
	}
	for(int i=0;i<MX;++i)	lst[i] = n+1;
	for(int i=n;i;--i)	nxt[i] = lst[a[i]] , lst[a[i]] = i;

	memset(tag,INF,sizeof tag);
	
	for(int i=1;i<=m;++i)	Q[i].l = read(),Q[i].r = read(),Q[i].id = i;
	
	sort(Q+1,Q+m+1);
	
	for(int i=1,now=1;i<=m;++i){
		while(now<Q[i].l){
			modify(1,1,n,now,nxt[now]-1,a[now]);
			++now;
		}
		ans[Q[i].id] = inqry(1,1,n,Q[i].r);
	}
	
	for(int i=1;i<=m;++i)	printf("%d\n",ans[i]);
	return 0;
}