problem
题意给定一棵树，并给定在这棵树上的两种操作。一种操作是改变一个节点的权值，另外一个操作是对两个节点之间的路径上的权值进行统计，如果每个权值出现的次数都是偶数，输出-1，否则输出出现次数为奇数的权值（保证只有一个）思路AC代码
g++开栈代码
register char *_sp __asm__(“rsp”);int main(){const int size = 64*1024*1024;[size]+size-4096);sys = _sp;_sp = mine;mmain();_sp = sys;return 0;};const int MAXN = 110000;int cnt,head[MAXN],dep[MAXN],p[MAXN][20],tot;int st[MAXN],ed[MAXN];int sum[MAXN<<3];int n,q;void update(int pos,int c,int l,int r,int rt){if(l == r){sum[rt] = c;return;}int m = (l+r)>>1;if(pos <= m)update(pos,c,lson);else update(pos,c,rson);sum[rt] = sum[rt<<1] ^ sum[rt<<1|1];}int query(int L,int R,int l,int r,int rt){if(l>=L && R>=r){return sum[rt];}int ret = 0;int m = (l+r)>>1;if(L <= m)ret ^= query(L,R,lson);if(R > m)ret ^= query(L,R,rson);return ret;}struct Edge{int u,v;int next;}e[MAXN<<1];void addedge(int u,int v){e[cnt].u = u,e[cnt].v = v,e[cnt].next = head[u],head[u] = cnt++;e[cnt].u = v,e[cnt].v = u,e[cnt].next = head[v],head[v] = cnt++;}void dfs(int u){st[u] = tot++;for(int i=head[u];i!=-1;i=e[i].next){int v = e[i].v;if(!dep[v]){dep[v] = dep[u] + 1;p[v][0] = u;dfs(v);}}ed[u] = tot++;}void init(){cnt = 0;tot = 1;memset(head,-1,sizeof(head));memset(p,-1,sizeof(p));memset(dep,0,sizeof(dep));}int LCA(int a,int b){if(dep[a]<dep[b])swap(a,b);int i;for(i = 0;(1<<i)<=dep[a];i++);i–;for(int j=i;j>=0;j–)if(dep[a]-(1<<j)>=dep[b])a = p[a][j];if(a == b)return a;for(int j=i;j>=0;j–){if(p[a][j]!=-1 && p[a][j] != p[b][j]){a = p[a][j];b = p[b][j];}}return p[a][0];}int u,v;int main(){int T;cin>>T;while(T–){scanf(“%d%d”,&n,&q);init();for(int i=0;i<n-1;i++){scanf(“%d%d”,&u,&v);addedge(u,v);}dep[1] = 1;dfs(1);for(int j=1;(1<<j)<=n;j++)for(int i=1;i<=n;i++)if(p[i][j-1] != -1)p[i][j] = p[p[i][j-1]][j-1];tot–;memset(sum,0,sizeof(sum));(int i=1;i<=n;i++){scanf(“%d”,&u);update(st[i],u+1,1,tot,1),update(ed[i],u+1,1,tot,1);}while(q–){int op,x,y;scanf(“%d%d%d”,&op,&x,&y);if(op == 0){update(st[x],y+1,1,tot,1),update(ed[x],y+1,1,tot,1);}else{int ansx = query(1,st[x],1,tot,1),ansy = query(1,st[y],1,tot,1);int lca = LCA(x,y),anslca =query(st[lca],st[lca],1,tot,1);//printf(“ansx:%d ansy:%d lca:%d\n”,ansx,ansy,lca);int ans = ansx ^ ansy ^ anslca;if( ans == 0)puts(“-1”);else printf(“%d\n”,ans-1);}}}return 0;}
，若不给自己设限，则人生中就没有限制你发挥的藩篱。
