# HDU 3887 counting offspring (Dfs sequence "non-recursive" + Tree-like array)

Source: Internet
Author: User

Test instructions

N points form a tree. gives the root node p and the tree structure information.

Outputs the f[i of each point]. F[i]: The number of numbers in all sub-nodes with a number smaller than I in the root of I.

0<n<=10^5

Ideas:

Method One: Direct DFS, enter the node x to record the number of smaller than X. When you come out X, record the number of smaller numbers than X. Subtraction is f[x]. Combines a tree-like array.

Method Two: Write down the DFS sequence. Builds a segment tree for the DFS sequence. Then the nodes are inserted from small to large. Statistics by line segment tree.

Code: (Method One)

`int Constn=1e5+5;intN,p;vector<int>G[n];intC[n];intAns[n];stack<int>S;BOOLVis[n];voidADD (intx) {     for(inti=x;i<=n;i+=lowbit (i)) {C[i]++; }}intQueryintx) {    intret=0;  for(intI=x;i>0; i-=lowbit (i)) {ret+=C[i]; }    returnret;}voidDfsintu) {     while(!S.empty ())    {S.pop (); } mem (Vis,false);    S.push (U); Vis[u]=true;  while(!S.empty ()) {        intnow=S.top (); intL=g[now].size (); BOOLflag=false; Rep (I,0, L-1){            if(vis[g[now][i]]==false) {Vis[g[now][i]]=true; Flag=true; intt=G[now][i]; Ans[t]=query (t1);                S.push (G[now][i]);  Break; }        }        if(!flag) {Ans[now]=query (now-1)-Ans[now];            ADD (now);        S.pop (); }    }}intMain () { while(SCANF ("%d%d", &n,&p)!=eof,n| |p) {Rep (i,1nn)        G[i].clear (); Mem (C,0); Mem (ans,0); Rep (I,1, N-1){            intb; scanf ("%d%d",&a,&b);            G[a].push_back (b);        G[b].push_back (a);        } DFS (P); Rep (I,1, N-1) printf ("#df", Ans[i]); cout<<ans[n]<<Endl; }    return 0;}`

HDU 3887 counting offspring (Dfs sequence "non-recursive" + Tree-like array)

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