hdu 6703 array(权值线段树)

Problem Description

You are given an array a1,a2,...,an(∀i∈[1,n],1≤ai≤n). Initially, each element of the array is **unique**.

Moreover, there are m instructions.

Each instruction is in one of the following two formats:

1. (1,pos),indicating to change the value of apos to apos+10,000,000;
2. (2,r,k),indicating to ask the minimum value which is **not equal** to any ai ( 1≤i≤r ) and **not less ** than k.

Please print all results of the instructions in format 2.

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, there are two integers n(1≤n≤100,000),m(1≤m≤100,000) in the first line, denoting the size of array a and the number of instructions.

In the second line, there are n distinct integers a1,a2,...,an (∀i∈[1,n],1≤ai≤n),denoting the array.
For the following m lines, each line is of format (1,t1) or (2,t2,t3).
The parameters of each instruction are generated by such way :

For instructions in format 1 , we defined pos=t1⊕LastAns . (It is promised that 1≤pos≤n)

For instructions in format 2 , we defined r=t2⊕LastAns,k=t3⊕LastAns. (It is promised that 1≤r≤n,1≤k≤n )

(Note that ⊕ means the bitwise XOR operator. )

Before the first instruction of each test case, LastAns is equal to 0 .After each instruction in format 2, LastAns will be changed to the result of that instruction.

(∑n≤510,000,∑m≤510,000 )

Output

For each instruction in format 2, output the answer in one line.

思路

有两种nlogn的解法 、

其一：用权值线段树维护出现的下标 然后维护一下区间最大值 对于1操作我们可以直接把下标变大 对于操作二我们可以区间查询到底 但是要加上一个判断 就是当前子树的

最大下标是否大于输入的位置。

其二：我们可以同样考虑用权值线段树维护当前的最左位置 这样建立可持久化线段树以后我们就可以解决不修改的查询操作 对于操作一 我们可以考虑用一个set维护 那么对于某一查询要么就是

直接查出来的值要么就是set里面的值 我们选一个最小的即可。

解法一：

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int N = 1e5+7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
typedef long long ll;
const ll mod = 1e9+7;
int a[N],po[N];
struct tree{
    int l,r,v;
}t[N<<2];
void build(int p,int l,int r){
    t[p].l=l; t[p].r=r;
    if(l==r){
        t[p].v=po[l];
        //cout<<t[p].l<<" "<<t[p].r<<" "<<t[p].v<<endl;
        return ;
    }
    int mid=(l+r)>>1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
    t[p].v=max(t[p<<1].v,t[p<<1|1].v);
}
void update(int p,int x,int v){
    if(t[p].l==t[p].r&&t[p].l==x){
        t[p].v=v;
        po[t[p].l]=v;
        return ;
    }
    int mid=(t[p].l+t[p].r)>>1;
    if(x<=mid) update(p<<1,x,v);
    else update(p<<1|1,x,v);
    t[p].v=max(t[p<<1].v,t[p<<1|1].v);
}
int query(int p,int l,int r,int pos){
//    cout<<t[p].l<<" "<<t[p].r<<endl;
    if(t[p].l==t[p].r){
        return t[p].l;
    }
    int mid=(t[p].l+t[p].r)>>1;
    int res;
    if(t[p<<1].v>pos){
        if(l<=mid) res=query(p<<1,l,r,pos);
        if(po[res]>pos) return res;
        if(r>mid) res=query(p<<1|1,l,r,pos);
    }else{
        if(r>mid) res=query(p<<1|1,l,r,pos);
    }
    return res;
}
int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0); cout.tie(0);
    int t; scanf("%d",&t);
    while(t--){
        memset(po,0,sizeof(po));
        int n,m; scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%d",a+i),po[a[i]]=i;
        po[n+1]=n+1;
        a[n+1]=n+1;
        build(1,1,n+1);
        int lastans=0,ans;
        for(int i=1;i<=m;i++){
            int op,r,k;
            scanf("%d%d",&op,&r);
            if(op==1){
                r=r^lastans;
                //printf("%d\n",a[r]);
                update(1,a[r],n+1);
            }else{
                scanf("%d",&k);
                r=r^lastans; k=k^lastans;
                //cout<<r<<" "<<k<<" "<<lastans<<endl;
                ans=query(1,k,n+1,r);
                lastans=ans;
                printf("%d\n",ans);
            //    cout<<ans<<"\n";
            }
        }
    }
} 

解法二：

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int N = 1e5+7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
typedef long long ll;
const ll mod = 1e7+9;
int a[N],rt[N];
struct tree{
    int l,r,v;
    int ls,rs;
}t[N<<5];
set<int> s;
int nico=0;
int cnt=0;
void build(int &p,int l,int r){
    p=++nico;
    t[p].l=l; t[p].r=r;
    if(l==r){
        t[p].v=l;
        return ;
    }
    int mid=(l+r)>>1;
    build(t[p].ls,l,mid);
    build(t[p].rs,mid+1,r);
    t[p].v=min(t[t[p].ls].v,t[t[p].rs].v);
}
void update(int &p,int last,int x,int v){
    p=++cnt;
    t[p]=t[last];
    if(t[p].l==t[p].r&&t[p].l==x){
        t[p].v=v;
        return ;
    }
    int mid=(t[p].l+t[p].r)>>1;
    if(x<=mid) update(t[p].ls,t[last].ls,x,v);
    else update(t[p].rs,t[last].rs,x,v);
    t[p].v=min(t[t[p].ls].v,t[t[p].rs].v);
}
int query(int p,int l,int r){
    if(l<=t[p].l&&t[p].r<=r){
        return t[p].v;
    }
    int mid=(t[p].l+t[p].r)>>1;
    int ans=inf;
    if(l<=mid) ans=min(ans,query(t[p].ls,l,r));
    if(mid<r) ans=min(ans,query(t[p].rs,l,r));
    return ans;
}
int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0); cout.tie(0);
    int t; scanf("%d",&t);

    while(t--){
        s.clear();
        int n,m; scanf("%d%d",&n,&m);
        build(rt[0],1,n+1);
        cnt=nico;
        for(int i=1;i<=n;i++){
            scanf("%d",a+i);
            update(rt[i],rt[i-1],a[i],inf);
        }
        int lastans=0; int nowans;
        for(int i=1;i<=m;i++){
            int op,t1,t2;
            scanf("%d",&op);
            if(op==1){
                scanf("%d",&t1);
                t1=t1^lastans;
                s.insert(a[t1]);
            }else{
                scanf("%d%d",&t1,&t2);
                t1=t1^lastans; t2=t2^lastans;
                auto v=s.lower_bound(t2);
                if(v!=s.end()){
                    nowans=min(*v,query(rt[t1],t2,n+1));
                }else{
                    nowans=query(rt[t1],t2,n+1);
                }
                lastans=nowans;
                printf("%d\n",nowans);
            }
        }
    }
}