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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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); } } } }