AC日记——Dylans loves tree hdu 5274

Dylans loves tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1611    Accepted Submission(s):
388

Problem Description

Dylans is given a tree with N

nodes.

All nodes have a value A[i]

.Nodes on tree is numbered by 1∼N

.

Then he is given Q

questions like that:

①0 x y

：change node x′s

value to y

②1 x y

：For all the value in the path from x

to y

,do they all appear even times?

For each ② question,it guarantees that
there is at most one value that appears odd times on the path.

1≤N,Q≤100000

, the value A[i]∈N

and A[i]≤100000

 

Input

In the first line there is a test number T

.
(T≤3

and there is at most one testcase that N>1000

)

For each testcase:

In the first line there are two numbers N

and Q

.

Then in the next N−1

lines there are pairs of (X,Y)

that stand for a road from x

to y

.

Then in the next line there are N

numbers A1..AN

stand for value.

In the next Q

lines there are three numbers(opt,x,y)

.

 

Output

For each question ② in each testcase,if the value all
appear even times output "-1",otherwise output the value that appears odd
times.

 

Sample Input

1
3 2
1 2
2 3
1 1 1
1 1 2
1 1 3

 

Sample Output

-1
1

Hint

If you want to hack someone,N and Q in your testdata must smaller than 10000，and you shouldn't print any space in each end of the line.

 

Source

BestCoder
Round #45

 

 

思路：

　　判断树上路径是否有出现过奇数次的权值；

　　数据保证一条路径上最多一个出现奇数次的权值；

　　因为x^x=0,0^x=x的性质；

　　我们可以用异或和；

　　把一条路径的全部权值都异或；

　　如果等于0则不存在，否则就输出；

　　同时，因为0^0=0；

　　所以我们全部的权值都+1；

　　然后，输出时再-1；

 

 

来，上代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

#define maxn 100005

using namespace std;

struct TreeNodeType {
    int l,r,dis,mid;
};
struct TreeNodeType tree[maxn<<];

struct EdgeType {
    int v,next;
};
struct EdgeType edge[maxn<<];

int if_z,t,n,q,dis[maxn],dis_[maxn];
int cnt=,head[maxn],deep[maxn],f[maxn];
int top[maxn],size[maxn],flag[maxn];

char Cget;

inline void in(int &now)
{
    now=,if_z=,Cget=getchar();
    while(Cget>''||Cget<'')
    {
        if(Cget=='-') if_z=-;
        Cget=getchar();
    }
    while(Cget>=''&&Cget<='')
    {
        now=now*+Cget-'';
        Cget=getchar();
    }
    now*=if_z;
}

inline void edge_add(int u,int v)
{
    cnt++;
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt;
}

void search_1(int now,int fa)
{
    int pos=cnt++;
    deep[now]=deep[fa]+,f[now]=fa;
    for(int i=head[now];i;i=edge[i].next)
    {
        if(edge[i].v==fa) continue;
        search_1(edge[i].v,now);
    }
    size[now]=cnt-pos;
}

void search_2(int now,int chain)
{
    top[now]=chain,flag[now]=++cnt;
    dis[flag[now]]=dis_[now]+;
    int pos=;
    for(int i=head[now];i;i=edge[i].next)
    {
        if(edge[i].v==f[now]) continue;
        if(size[edge[i].v]>size[pos]) pos=edge[i].v;
    }
    if(pos==) return ;
    search_2(pos,chain);
    for(int i=head[now];i;i=edge[i].next)
    {
        if(edge[i].v==pos||edge[i].v==f[now]) continue;
        search_2(edge[i].v,edge[i].v);
    }
}

inline void tree_up(int now)
{
    tree[now].dis=tree[now<<].dis^tree[now<<|].dis;
}

void tree_build(int now,int l,int r)
{
    tree[now].l=l,tree[now].r=r;
    if(l==r)
    {
        tree[now].dis=dis[l];
        return ;
    }
    tree[now].mid=(l+r)>>;
    tree_build(now<<,l,tree[now].mid);
    tree_build(now<<|,tree[now].mid+,r);
    tree_up(now);
}

void tree_change(int now,int to,int x)
{
    if(tree[now].l==tree[now].r)
    {
        tree[now].dis=x;
        return ;
    }
    if(to<=tree[now].mid) tree_change(now<<,to,x);
    else tree_change(now<<|,to,x);
    tree_up(now);
}

int tree_query(int now,int l,int r)
{
    if(tree[now].l==l&&tree[now].r==r)
    {
        return tree[now].dis;
    }
    if(l>tree[now].mid) return tree_query(now<<|,l,r);
    else if(r<=tree[now].mid) return tree_query(now<<,l,r);
    else return tree_query(now<<,l,tree[now].mid)^tree_query(now<<|,tree[now].mid+,r);
}

int solve_do(int x,int y)
{
    int pos=;
    while(top[x]!=top[y])
    {
        if(deep[top[x]]<deep[top[y]]) swap(x,y);
        pos=pos^tree_query(,flag[top[x]],flag[x]);
        x=f[top[x]];
    }
    if(deep[x]>deep[y]) swap(x,y);
    pos=pos^tree_query(,flag[x],flag[y]);
    if(pos==) return -;
    else return pos-;
}

int main()
{
    in(t);
    int lit=t;
    while(t--)
    {
        memset(head,,sizeof(head));
        in(n),in(q);cnt=;int u,v;
        for(int i=;i<n;i++)
        {
            in(u),in(v);
            edge_add(u,v);
            edge_add(v,u);
        }
        for(int i=;i<=n;i++) in(dis_[i]);
        cnt=,search_1(,);
        cnt=,search_2(,);
        tree_build(,,n);
        int type;
        for(int i=;i<=q;i++)
        {
            in(type),in(u),in(v);
            if(type) printf("%d\n",solve_do(u,v));
            else tree_change(,u,v+);
        }
    }
    return ;
}