[PKU 1330]Nearest Common Ancestors（LCA）

 

【题目大意】

给你一棵树，让你找两个点的最近公共祖先。

【题目分析】：

裸的LCA，实际上可以用朴素的骗过。但是为了学习伟大的LCA离线Tarjan算法，还是使用了它~

LCA讲解详见：http://hi.baidu.com/niuniu_2006923/blog/item/3d57d0235ce7ae5a9922ed68.html

【代码】：

program PKU_1330;type ntype=record data,next:longint; end;var g:array[1..10000] of longint; node:array[1..10000] of ntype; sign:array[1..10000] of boolean; father:array[1..10000] of longint; t,n,tt,i,tx,ty,x,y,p,tot,root:longint; flag:boolean;function find(x:longint):longint;begin if father[x]=x then exit(x); father[x]:=find(father[x]); exit(father[x]);end;procedure lca(x:longint);var p:longint;begin if flag then exit; father[x]:=x; if (tx=x) and (father[ty]<>0) then begin writeln(find(ty)); flag:=true; exit; end; if (ty=x) and (father[tx]<>0) then begin writeln(find(tx)); flag:=true; exit; end; p:=g[x]; while p<>0 do begin lca(node[p].data); father[node[p].data]:=find(x); p:=node[p].next; end;end;begin assign(input,'a.in'); assign(output,'a.out'); reset(input); rewrite(output); readln(t); for tt:=1 to t do begin readln(n); tot:=0; fillchar(g,sizeof(g),0); fillchar(node,sizeof(node),0); fillchar(sign,sizeof(sign),0); for i:=1 to n-1 do begin readln(x,y); inc(tot); p:=tot; node[p].data:=y; node[p].next:=g[x]; g[x]:=p; sign[y]:=true; end; for i:=1 to n do if not sign[i] then begin root:=i; break; end; readln(tx,ty); flag:=false; fillchar(father,sizeof(father),0); lca(root); end; close(input); close(output);end.