题目链接：
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1028
比较简单，用数组d[i]表示结点i和祖先结点的距离，查询时压缩路径并更新d[]数组。
刘汝佳大白书（P193）：
贴代码：
1 #include<cstdio> 2 #include<algorithm> 3 const int N = 20004; 4 int pa[N],d[N]; 5 int findset(int x) 6 { 7 if(pa[x] == -1) return x; 8 else 9 { 10 int root = findset(pa[x]); 11 d[x] += d[pa[x]]; 12 return pa[x] = root; 13 } 14 } 15 int main() 16 { 17 // freopen("in.txt","r",stdin); 18 int t,n,u,v; 19 scanf("%d",&t); 20 while(t--) 21 { 22 scanf("%d",&n); 23 char s[4]; 24 for(int i=0; i<=n; ++i) pa[i]=-1,d[i]=0; 25 while(scanf("%s",s),s[0] != 'O') 26 { 27 if(s[0] == 'E') 28 { 29 scanf("%d",&u); 30 findset(u); 31 printf("%d ",d[u]); 32 } 33 else 34 { 35 scanf("%d%d",&u,&v); 36 pa[u] = v; 37 d[u] = abs(u-v)%1000; 38 } 39 } 40 } 41 return 0; 42 }