[BZOJ3585]mex（离线+离散化+线段树）

来源：互联网发布：蒙大拿州立大学知乎编辑：程序博客网时间：2024/05/06 15:24

题目描述

传送门

题解

和BZOJ3339基本一样。但是需要离散化求next，然后求sg我用了map。

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<map>using namespace std;const int max_n=2e5+5;const int max_tree=max_n*4;const int INF=2e9;int n,m,cnt,k,L,a[max_n],b[max_n],p[max_n],sg[max_n],next[max_n],last[max_n];struct hp{int l,r,id,ans;}q[max_n];int tree[max_tree],delta[max_tree];map <int,bool>mark;inline int cmp1(int x,int y){return a[x]<a[y];}inline int cmp(hp a,hp b){return a.l<b.l;}inline void update(int now){tree[now]=min(tree[now<<1],tree[now<<1|1]);}inline void pushdown(int now,int l,int r,int mid){    if (delta[now]!=INF){        tree[now<<1]=min(tree[now<<1],delta[now]);        delta[now<<1]=min(delta[now<<1],delta[now]);        tree[now<<1|1]=min(tree[now<<1|1],delta[now]);        delta[now<<1|1]=min(delta[now<<1|1],delta[now]);        delta[now]=INF;    }}inline void build(int now,int l,int r){    int mid=(l+r)>>1;delta[now]=INF,tree[now]=INF;    if (l==r){        tree[now]=sg[l];        return;    }    build(now<<1,l,mid);    build(now<<1|1,mid+1,r);    update(now);}inline void interval_change(int now,int l,int r,int lrange,int rrange,int v){    int mid=(l+r)>>1;    if (lrange<=l&&r<=rrange){        tree[now]=min(tree[now],v);        delta[now]=min(delta[now],v);        return;    }    pushdown(now,l,r,mid);    if (lrange<=mid) interval_change(now<<1,l,mid,lrange,rrange,v);    if (mid+1<=rrange) interval_change(now<<1|1,mid+1,r,lrange,rrange,v);    update(now);}inline int query(int now,int l,int r,int x){    int mid=(l+r)>>1;    if (l==r) return tree[now];    pushdown(now,l,r,mid);    if (x<=mid) return query(now<<1,l,mid,x);    else return query(now<<1|1,mid+1,r,x);}int main(){    scanf("%d%d",&n,&m);    for (int i=1;i<=n;++i) scanf("%d",&a[i]),p[i]=i;    sort(p+1,p+n+1,cmp1);    for (int i=1;i<=n;++i)      if (a[p[i]]==a[p[i-1]]) b[p[i]]=cnt;      else b[p[i]]=++cnt;    k=0;    for (int i=1;i<=n;++i){        mark[a[i]]=1;        while (mark[k]) k++;        sg[i]=k;    }    for (int i=1;i<=n;++i) next[i]=n+1;    for (int i=1;i<=n;++i){        next[last[b[i]]]=i;        last[b[i]]=i;    }    build(1,1,n);    for (int i=1;i<=m;++i)      scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i;    sort(q+1,q+m+1,cmp);    L=1;    for (int i=1;i<=m;++i){        while (L<q[i].l){            interval_change(1,1,n,L,next[L]-1,a[L]);            L++;        }        q[q[i].id].ans=query(1,1,n,q[i].r);    }    for (int i=1;i<=m;++i)      printf("%d\n",q[i].ans);}

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