Corporative Network (with right and check set)

The test instructions of this question is when the input ' E ' is the lookup operation, looking for the value from the back of this number to his father side, ' I ' represents the father who put the number of the back as the preceding number

Then the difference between them represents the weights of the two sides.

Water and water problems

#include <stdio.h> #include <string.h>int par[20005];int rank1[20005];int abs (int hh) {return (hh>0)? HH: -HH;}    void Init () {for (int i=0;i<20005;i++) {par[i]=i;    rank1[i]=0; }}int findset (int x) {int root;if (x!=par[x]) {root=findset (par[x]); Rank1[x]+=rank1[par[x]];p ar[x]=root;} return par[x];}            int main () {int t;int k;int A,u,v;char s[10];while (scanf ("%d", &t) ==1) {while (t--) {scanf ("%d", &k);          Init ();          while (scanf ("%s", s)!=eof) {if (s[0]== ' O ') break;          if (s[0]== ' E ') {scanf ("%d", &a);          Findset (a);          printf ("%d\n", Rank1[a]);          } else {scanf ("%d%d", &u,&v);          Par[u]=v;          Rank1[u]=abs (u-v)%1000;          printf ("%d%d\n", U,rank1[u]);          }//printf ("%d%d%d%d\n", rank1[1],rank1[2],rank1[3],rank1[4]); }}}}