POJ 1330 Nearest Common Ancestors (template title) (LCA) "multiplier"

< topic links >

Main topic:

Give a tree and ask the number of the nearest public ancestor of any two points.

Problem Solving Analysis:
LCA template problem, the following is the online multiplication algorithm solution.

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <cmath>5 using namespacestd;6 7 Const intN = 1e4+Ten;8 Const intINF =0x3f3f3f3f;9 structedge{Ten     intTo,next; One}edge[n<<1]; A intCnt,head[n]; - intdep[n],f[n][ *]; - intNinch[N]; the voidAddedge (intUintv) { -edge[++cnt].to=v,edge[cnt].next=Head[u]; -head[u]=CNT; - } + voidDfsintUintFA) {//mark the depth of all nodes -      for(intI=head[u];~i;i=Edge[i].next) { +         intv=edge[i].to; A         if(V==FA)Continue; at         if(!Dep[v]) { -dep[v]=dep[u]+1; -f[v][0]=u; - DFS (v,u); -         }      -     } in } - voidInit () {//preprocessing multiplication Arrays to      for(intj=1;(1&LT;&LT;J) <=n;j++) +          for(intI=1; i<=n;i++) -f[i][j]=f[f[i][j-1]][j-1]; the } * intLCA (intXinty) { $     if(dep[x]<Dep[y]) swap (x, y);Panax Notoginseng     intd=dep[x]-Dep[y]; -      for(intI=0;(d >>i)! =0; i++) the         if((d>>i) &1) x=F[x][i]; +     if(x==y)returnx; A      for(intI= -; i>=0; i--) the         if(f[x][i]!=F[y][i]) { +x=F[x][i]; -y=F[y][i]; $         } $     returnf[x][0]; - } - intMain () { the     intT;SCANF ("%d",&T); -      while(t--){Wuyiscanf"%d",&n); theCnt=0; -         intu,v; Wumemset (head,-1,sizeof(head)); -          for(intI=1; i<n;i++){             Aboutscanf"%d%d",&u,&v); $ Addedge (u,v); -             inch[v]++; -         } -memset (DEP,0,sizeof(DEP)); A         intRoot; +          for(intI=1; i<=n;i++) the             if(!inch[i]) root=i;//find the root of the tree -dep[root]=1; $DFS (root,-1); the init (); thescanf"%d%d",&u,&v); theprintf"%d\n", LCA (u,v)); the     } -     return 0; in}

2018-10-18

POJ 1330 Nearest Common Ancestors (template title) (LCA) "multiplier"