Dylans loves tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1915 Accepted Submission(s): 492

Problem Description

Dylans is given a tree with

Input

In the first line there is a test number

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.

思路：

一道非常简单的题。。要求u-v之前出现次数为奇数的数字，如果没找到就输出-1，找到就输出这个数字。

题目保证了出现为奇数次的数字的个数不超过1，那么直接对u-v异或就好了，最后剩下的如果是0有两种可能：

1.没有出现次数为奇数的数字。 2.出现次数为奇数的数字为0；

我们只要将输入的数字全部+1就可以避免这种情况了

最后当输出的值为0时没有出现次数为奇数的数字，输出-1，不为0的话直接输出这个数字就好了。

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mid int m = (l + r) >> 1
const int M = 2e5+10;
struct node{
    int to,next;
}e[M];
int sum[M<<2],son[M],fa[M],head[M],siz[M],top[M],dep[M],tid[M],rk[M],a[M];
int cnt1,cnt,n;
void add(int u,int v){
    e[++cnt1].to = v;e[cnt1].next = head[u];head[u] = cnt1;
    e[++cnt1].to = u;e[cnt1].next = head[v];head[v] = cnt1;
}

void dfs1(int u,int faz,int deep){
     dep[u] = deep;
     fa[u] = faz;
     siz[u] = 1;
     for(int i = head[u];i;i=e[i].next){
        int v = e[i].to;
        if(v != fa[u]){
            dfs1(v,u,deep+1);
            siz[u] += siz[v];
            if(son[u] == -1||siz[v] > siz[son[u]])
                son[u] = v;
        }
     }
}

void dfs2(int u,int t){
    top[u] = t;
    tid[u] = cnt;
    rk[cnt] = u;
    cnt++;
    if(son[u] == -1) return;
    dfs2(son[u],t);
    for(int i = head[u];i;i = e[i].next){
        int v = e[i].to;
        if(v != son[u]&&v != fa[u])
            dfs2(v,v);
    }
}

void pushup(int rt){
    sum[rt] = sum[rt<<1]^sum[rt<<1|1];
}

void build(int l,int r,int rt){
     if(l == r){
        sum[rt] = a[rk[l]];
        return ;
     }
     mid;
     build(lson);
     build(rson);
     pushup(rt);
}

void update(int p,int c,int l,int r,int rt){
     if(l == r){
        sum[rt] = c;
        return ;
     }
     mid;
     if(p <= m) update(p,c,lson);
     else update(p,c,rson);
     pushup(rt);
}

int query(int L,int R,int l,int r,int rt){
    if(L <= l&&R >= r){
        return sum[rt];
    }
    mid;
    int ret = 0;
    if(L <= m) ret^=query(L,R,lson);
    if(R > m) ret^=query(L,R,rson);
    return ret;
}

int ask(int x,int y){
    int sum = 0;
    int fx = top[x],fy = top[y];
    while(fx != fy){
        if(dep[fx] < dep[fy]) swap(x,y),swap(fx,fy);
        sum ^= query(tid[fx],tid[x],1,n,1);
        x = fa[fx];fx = top[x];
    }
    if(dep[x] < dep[y]) swap(x,y);
    sum^=query(tid[y],tid[x],1,n,1);
    //cout<<sum<<endl;
    return sum;
}

void init()
{
    memset(son,-1,sizeof(son));
    for(int i = 0; i <= 2*n;i ++){
    e[i].to = 0;e[i].next = 0;head[i] = 0;
    }
}
int main()
{
    int t,u,v,x,y,op,q;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&q);
        cnt = 1; cnt1 = 1;
        init();
        for(int i = 0; i < n-1;i ++){
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        for(int i = 1;i <= n;i ++)
            scanf("%d",&x),a[i]=x+1;
        dfs1(1,0,1); dfs2(1,1); build(1,n,1);
        while(q--){
            scanf("%d",&op);
            if(op==1){
                scanf("%d%d",&x,&y);
                //cout<<ask(x,y)<<endl;
                if(ask(x,y)!=0) printf("%d\n",ask(x,y)-1);
                else printf("-1\n");
            }
            else{
                scanf("%d%d",&x,&y);
                update(tid[x],y+1,1,n,1);
            }
        }
    }
    return 0;
}

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