"Algorithmic Learning Notes" 91. Simple forest count SJTU OJ 10,452 Brother's family

In fact, huge water ... But no scanf prinf at all. It's boring.

The first version of the code is a little bit inefficient, mainly considering the possibility that the family might have started out as a forest, not from a tree.

Actually, the data point is still a tree, and then it becomes a forest. So, as long as three arrays are available.

Alive is the record alive?

Sons record the number of child nodes for each person

Father records each person's parent can determine if x is the root node of a tree according to Alive[father[x]].

CNT Maintenance Family Number

Every time a dead person

Judging whether a dead person is a root node of a tree or just a leaf node.

#include <iostream>#include<vector>#include<algorithm>#include<cstdio>using namespacestd;Const intMAXN =200000+Ten;intN;intFATHER[MAXN] = {0};//record who the father of each person isBOOLALIVE[MAXN] = {false};//is the record alive ?intsons[maxn]={0};//record the number of sonsvoidinit () {cin>>N;}intCNT =1;//Number of familiesvoidbuild () { for(inti =0; i < N; ++i) {        Charop; Op=GetChar ();  while(op!='B'and op!='D') {op=GetChar (); }         while(OP! ='B'&& OP! ='D') scanf ("%c",&op); intx, y; Switch(OP) { Case 'B': scanf ("%d%d",&x,&y); if(!alive[x]) sons[x]=0; Father[y]= x;//Y's dad's X.ALIVE[X] = Alive[y] =true;//They 're both alive.sons[x]++;  Break;  Case 'D': scanf ("%d",&x); ALIVE[X]=false; if(!alive[father[x]])//It 's a root node, Dad's dead.cnt--; ElseSons[father[x]]--; CNT+=Sons[x]; //cout<<cnt<<endl;printf"%d\n", CNT);  Break; }    }}intMainintargcChar Const*argv[])    {init ();    Build (); return 0;}

"Algorithmic Learning Notes" 91. Simple forest count SJTU OJ 10,452 Brother's family