array(2019CCPC网络预选赛 hdu6703主席树+set)主席树求大于等于k的最小值

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.

Sample Input
3
5 9
4 3 1 2 5
2 1 1
2 2 2
2 6 7
2 1 3
2 6 3
2 0 4
1 5
2 3 7
2 4 3
10 6
1 2 4 6 3 5 9 10 7 8
2 7 2
1 2
2 0 5
2 11 10
1 3
2 3 2
10 10
9 7 5 3 4 10 6 2 1 8
1 10
2 8 9
1 12
2 15 15
1 12
2 1 3
1 9
1 12
2 2 2
1 9

Sample Output
1
5
2
2
5
6
1
6
7
3
11
10
11
4
8
11

Hint

note:
After the generation procedure ,the instructions of the first test case are :
2 1 1, in format 2 and r=1 , k=1
2 3 3, in format 2 and r=3 , k=3
2 3 2, in format 2 and r=3 , k=2
2 3 1, in format 2 and r=3 , k=1
2 4 1, in format 2 and r=4 , k=1
2 5 1, in format 2 and r=5 , k=1
1 3 , in format 1 and pos=3
2 5 1, in format 2 and r=5 , k=1
2 5 2, in format 2 and r=5 , k=2

the instructions of the second test case are :
2 7 2, in format 2 and r=7 , k=2
1 5 , in format 1 and pos=5
2 7 2, in format 2 and r=7 , k=2
2 8 9, in format 2 and r=8 , k=9
1 8 , in format 1 and pos=8
2 8 9, in format 2 and r=8 , k=9

the instructions of the third test case are :
1 10 , in format 1 and pos=10
2 8 9 , in format 2 and r=8 , k=9
1 7 , in format 1 and pos=7
2 4 4 , in format 2 and r=4 , k=4
1 8 , in format 1 and pos=8
2 5 7 , in format 2 and r=5 , k=7
1 1 , in format 1 and pos=1
1 4 , in format 1 and pos=4
2 10 10, in format 2 and r=10 , k=10
1 2 , in format 1 and pos=2
昨天卡了很久的一个题,你开始线段树+set一直超时,后来改了主席树,但是还是卡了很久。。
题意:要找出一个数,这个数在1~r中不能出现过,并且还要不小于k。并且读入的数据还都要异或上上一次的答案。
一开始理解错了,以为必须是数组里的数才行。但是第一个样例中就有一个数不是数组里的数,除此之外,题目中有说所有的数都是独一无二的。在执行操作1的时候,一个数加上1e7之后,就相当于作废了。因为数组中所有的数字都是不大于1e5的,所以异或之后的值也是不大于1e5的。这样的话,在加上1e7之后,就相当于这个值在1~r中删除了,又可以再用了。那这样我们把这样的值存到一个set里面,r+1到n的值用主席树去求。主席树是好多权值线段树的集合,是按着权值来的,存的是每个数的出现的个数。用主席树去求r+1 ~ n中大于等于k最小的数。
代码如下:

#include
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;

const int maxx=1e5+100;
struct node{
     
	int l;
	int r;
	int sum;
}p[maxx<<5];
int a[maxx];
int root[maxx];
int n,m,tot;

inline void pushup(int cur)
{
     
	p[cur].sum=p[p[cur].l].sum+p[p[cur].r].sum;
}
inline int build(int l,int r)
{
     
	int cur=++tot;
	p[cur].sum=0;
	if(l==r) return cur;
	int mid=l+r>>1;
	build(l,mid);
	build(mid+1,r);
	return cur;
}
inline int update(int rot,int l,int r,int pos)
{
     
	int cur=++tot;
	p[cur]=p[rot];
	if(l==r) 
	{
     
		p[cur].sum++;
		return cur;
	}
	int mid=l+r>>1;
	if(pos<=mid) p[cur].l=update(p[rot].l,l,mid,pos);
	else p[cur].r=update(p[rot].r,mid+1,r,pos);
	pushup(cur);
	return cur;
}
inline int query(int lrot,int rrot,int l,int r,int k)
{
     
	if(l==r) //到达叶子节点后
	{
     
		if(p[rrot].sum-p[lrot].sum>0) return l;//如果这个值出现过就返回这个值,没有就返回无穷大
		else return inf;
	}
	int mid=l+r>>1;
	if(k<=mid)
	{
     
		int ans=inf;
		if(k<=l)//这里相当于一个剪枝,在小于l的时候,如果左子树有符合条件的就进入左子树,否则再进入右子树。
		{
     
			if(p[p[rrot].l].sum-p[p[lrot].l].sum>0) ans=min(ans,query(p[lrot].l,p[rrot].l,l,mid,k));
			else if(p[p[rrot].r].sum-p[p[lrot].r].sum>0) ans=min(ans,query(p[lrot].r,p[rrot].r,mid+1,r,k));
			return ans;
		}
		if(p[p[rrot].l].sum-p[p[lrot].l].sum>0) //k在l到mid之间的时候,左右子树都有可能涉及,就左右都看一遍寻找最优解
		ans=min(ans,query(p[lrot].l,p[rrot].l,l,mid,k));
		if(p[p[rrot].r].sum-p[p[lrot].r].sum>0) ans=min(ans,query(p[lrot].r,p[rrot].r,mid+1,r,k));
		return ans;
	}
	else
	{
     
		int ans=inf;//k大于mid的时候,直接进入右子树,左子树不用找了
		if(p[p[rrot].r].sum-p[p[lrot].r].sum>0) ans=min(ans,query(p[lrot].r,p[rrot].r,mid+1,r,k));
		return ans;
	}
}

int main()
{
     
	int t,t1,t2,t3,op;
	scanf("%d",&t);
	ll lastans=0;
	while(t--)
	{
     
		scanf("%d%d",&n,&m);
		set<ll> s;
		s.insert(n+1);//一开始先把n+1放入集合,如果数组中的值没有符合的就输出n+1
		lastans=0;tot=0;
		for(int i=1;i<=n;i++) scanf("%d",&a[i]);
		root[0]=build(1,100001);
		for(int i=1;i<=n;i++) root[i]=update(root[i-1],1,100001,a[i]);
		while(m--)
		{
     
			scanf("%d",&op);
			if(op==1)
			{
     
				scanf("%d",&t1);
				t1^=lastans;
				s.insert(a[t1]);//把修改之后的值放入集合里
			}
			else if(op==2)
			{
     
				scanf("%d%d",&t2,&t3);
				t2^=lastans,t3^=lastans;
				int ans=inf;
				if(t2!=n) ans=query(root[t2],root[n],1,100001,t3);
				int ans1=*s.lower_bound(t3);
				lastans=min(ans1,ans);
				printf("%d\n",lastans);
			}
		}
	}
	return 0;
}

昨天写的超时的代码(线段树+set)当个借鉴吧
代码如下:

#include
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;

const int maxx=1e5+100;
struct node{
     
	int l;
	int r;
	set<int> s;
}p[maxx<<2];
int a[maxx];
int n,m;

inline int read()     //输入外挂
{
     
	int res=0,ch,flag=0;
	if((ch=getchar())=='-')
		flag=1;
	else if(ch>='0'&&ch<='9')
		res=ch-'0';
	while((ch=getchar())>='0'&&ch<='9')
		res=res*10+ch-'0';
	return flag?-res:res;
}
inline void Out(int a)    //输出外挂
{
     
	if(a>9)
		Out(a/10);
	putchar(a%10+'0');
}
inline void pushup(int cur)
{
     
	set_union(p[cur<<1].s.begin(),p[cur<<1].s.end(),p[cur<<1|1].s.begin(),p[cur<<1|1].s.end(),inserter(p[cur].s,p[cur].s.begin()));
}
inline void build(int l,int r,int cur)
{
     
	p[cur].l=l;
	p[cur].r=r;
	p[cur].s.clear();
	if(l==r) 
	{
     
		p[cur].s.insert(a[l]);
		return ;
	}
	int mid=l+r>>1;
	build(l,mid,cur<<1);
	build(mid+1,r,cur<<1|1);
	pushup(cur);
}
inline set<int> query(int l,int r,int cur)
{
     
	int L=p[cur].l;
	int R=p[cur].r;
	if(l<=L&&R<=r) return p[cur].s;
	int mid=L+R>>1;
	if(r<=mid) return query(l,r,cur<<1);
	else if(l>mid) return query(l,r,cur<<1|1);
	else
	{
     
		set<int> a=query(l,mid,cur<<1);
		set<int> b=query(mid+1,r,cur<<1|1);
		set<int> x;
		set_union(a.begin(),a.end(),b.begin(),b.end(),inserter(x,x.begin()));
		return x;
	}
}
int main()
{
     
	int t,t1,t2,t3,op;
	t=read();
	while(t--)
	{
     
		int lastans=0;
		n=read(),m=read();
		for(int i=1;i<=n;i++) a[i]=read();
		build(1,n,1);
		set<int> s;
		s.insert(n+1);
		while(m--)
		{
     
			op=read();
			if(op==1)
			{
     
				t1=read();
				t1^=lastans;
				s.insert(a[t1]);
			}
			else
			{
     
				t2=read(),t3=read();
				t2^=lastans;t3^=lastans;
				int ans1=inf;
				if(t2!=n) 
				{
     
					set<int> s1;
					s1=query(t2+1,n,1);
					s1.insert(n+1);
					ans1=*s1.lower_bound(t3);
				}
				int ans2=*s.lower_bound(t3);
				lastans=min(ans1,ans2);
				Out(lastans);
			}
		}
	}
}

努力加油a啊,(o)/~

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