Tree
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1643 Accepted Submission(s): 461
Problem Description
Zero and One are good friends who always have fun with each other. This time, they decide to do something on a tree which is a kind of graph that there is only one path from node to node. First, Zero will give One an tree and every node in this tree has a value. Then, Zero will ask One a series of queries. Each query contains three parameters: x, y, z which mean that he want to know the maximum value produced by z xor each value on the path from node x to node y (include node x, node y). Unfortunately, One has no idea in this question. So he need you to solve it.
Input
There are several test cases and the cases end with EOF. For each case:
The first line contains two integers n(1<=n<=10^5) and m(1<=m<=10^5), which are the amount of tree’s nodes and queries, respectively.
The second line contains n integers a[1..n] and a[i](0<=a[i]<2^{16}) is the value on the ith node.
The next n–1 lines contains two integers u v, which means there is an connection between u and v.
The next m lines contains three integers x y z, which are the parameters of Zero’s query.
Output
For each query, output the answer.
Sample Input
3 2
1 2 2
1 2
2 3
1 3 1
2 3 2
Sample Output
3
题目链接：HDU 4757
一道跟COT很像的题，但是用的是可持久化的Trie，做法跟COT基本相同，但是过程中由于N少设了10倍，无限TLE……各种纠结问题出在Tarjan？并查集？插入查询函数？因此过程中还找了题解但发现并没有什么区别，但有另外一种非递归形式的插入查询写法，值得借鉴。另外处理节点LCA的问题上公式还是那条$cnt_{U,V}=cnt_{U}+cnt_{V}-cnt_{LCA}-cnt_{father[LCA]}$
单纯地减掉两倍可能会出错，不知道网上的例程是什么情况。通过最近几道题了解了可持久化Trie的空间复杂度似乎跟Trie是一样的，都是Total*maxlen（如果撇开相对来说只有一丢丢的root数组不说的话）
递归更新查询的代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=1e5+7;
struct Trie
{
int nxt[2];
int cnt;
};
struct edge
{
int to,nxt;
};
struct Query
{
int to,nxt,id,lca;
};
struct ask
{
int u,v,lca,val;
};
ask q[N];
Trie L[N*18];int cnt;
Query Q[N<<1];int qtot,qhead[N];
edge E[N<<1];int tot,head[N];
int ances[N],pre[N],Father[N];
bitset<N>vis;
int arr[N],root[N]; void init()
{
CLR(L,0);cnt=0;
CLR(head,-1);tot=0;
CLR(qhead,-1);qtot=0;
CLR(ances,0);
for (int i=0; i<N; ++i)
pre[i]=i;
CLR(Father,0);
vis.reset();
CLR(root,0);
}
inline void add(const int &s,const int &t)
{
E[tot].to=t;
E[tot].nxt=head[s];
head[s]=tot++;
}
inline void addQ(const int &s,const int &t,const int &id)
{
Q[qtot].to=t;
Q[qtot].id=id;
Q[qtot].lca=1;
Q[qtot].nxt=qhead[s];
qhead[s]=qtot++;
} void update(int &cur,const int &ori,const int &step,const int &n)
{
cur=++cnt;
L[cur]=L[ori];
++L[cur].cnt;
if(step<0)
return;
int t=(n>>step)&1;
update(L[cur].nxt[t],L[ori].nxt[t],step-1,n);
}
int Find(const int &n)
{
return n==pre[n]?n:pre[n]=Find(pre[n]);
}
void Tarjan(const int &u,const int &Fa)
{
vis[u]=1;
ances[u]=u;
Father[u]=Fa;
update(root[u],root[Fa],16,arr[u]);
for (int i=head[u]; ~i; i=E[i].nxt)
{
int v=E[i].to;
if(!vis[v])
{
Tarjan(v,u);
pre[v]=u;
ances[Find(u)]=u;
}
}
for (int i=qhead[u]; ~i; i=Q[i].nxt)
{
int v=Q[i].to;
if(vis[v])
q[Q[i].id].lca=ances[Find(v)];
}
}
int query(const int &U,const int &V,const int &LCA,const int &F_LCA,const int &step,const int &n)
{
if(step<0)
return 0;
int t=(n>>step)&1;
int c=L[L[U].nxt[t^1]].cnt+L[L[V].nxt[t^1]].cnt-L[L[LCA].nxt[t^1]].cnt-L[L[F_LCA].nxt[t^1]].cnt;
if(c>0)
return (1<<step)+query(L[U].nxt[t^1],L[V].nxt[t^1],L[LCA].nxt[t^1],L[F_LCA].nxt[t^1],step-1,n);
else
return query(L[U].nxt[t],L[V].nxt[t],L[LCA].nxt[t],L[F_LCA].nxt[t],step-1,n);
}
int main(void)
{
int n,m,a,b,i;
while (~scanf("%d%d",&n,&m))
{
init();
for (i=1; i<=n; ++i)
scanf("%d",&arr[i]);
for (i=0; i<n-1; ++i)
{
scanf("%d%d",&a,&b);
add(a, b);
add(b, a);
}
for (i=0; i<m; ++i)
{
scanf("%d%d%d",&q[i].u,&q[i].v,&q[i].val);
addQ(q[i].u, q[i].v, i);
addQ(q[i].v, q[i].u, i);
}
Tarjan(1,0);
for (i=0; i<m; ++i)
printf("%d\n",query(root[q[i].u],root[q[i].v],root[q[i].lca],root[Father[q[i].lca]],16,q[i].val));
}
return 0;
}
网上例程的非递归写法：
int update(int ori,int n)
{
bitset<17> s=n;
int cur=++cnt;
int ret=cur;
L[cur]=L[ori];
for (int i=16; i>=0; --i)
{
int v=s[i];
L[++cnt]=L[L[cur].nxt[v]];
++L[cnt].cnt;
L[cur].nxt[v]=cnt;
cur=cnt;
}
return ret;
}
void Tarjan(const int &u,const int &Fa)
{
/*vis[u]=1;
ances[u]=u;
Father[u]=Fa;*/
root[u]=update(root[Fa],arr[u]);
/*for (int i=head[u]; ~i; i=E[i].nxt)
{
int v=E[i].to;
if(!vis[v])
{
Tarjan(v,u);
pre[v]=u;
ances[Find(u)]=u;
}
}
for (int i=qhead[u]; ~i; i=Q[i].nxt)
{
int v=Q[i].to;
if(vis[v])
q[Q[i].id].lca=ances[Find(v)];
}*/
}
int query(int u,int v,int lca,int flca,int n)
{
int r=0;
bitset<17> s=n;
for (int i=16; i>=0; --i)
{
int indx=s[i];
int c=L[L[u].nxt[indx^1]].cnt+L[L[v].nxt[indx^1]].cnt-L[L[lca].nxt[indx^1]].cnt-L[L[flca].nxt[indx^1]].cnt;
if(c>0)
{
r+=1<<i;
u=L[u].nxt[indx^1];
v=L[v].nxt[indx^1];
lca=L[lca].nxt[indx^1];
flca=L[flca].nxt[indx^1];
}
else
{
u=L[u].nxt[indx];
v=L[v].nxt[indx];
lca=L[lca].nxt[indx];
flca=L[flca].nxt[indx];
}
}
return r;
}