POJ1330&&LCA最近公用祖先模板

在這我用的是tarjan搞的：tarjan的話需要知道樹的根節點，這題沒說，但是可以根據入度等於0，判斷根節點;tarjan的主要思想是DFS和並查集;

tarjan:有待補充

#include<cstdio>#include<string.h>#include<algorithm>#include<vector>#include<stdlib.h>#include<cmath>#include<queue>#include<set>#include<map>#include<list>using namespace std;const int N=10005;vector<int>M[N],Q[N];bool vis[N];int father[N],ance[N],in[N];int ex,ey;int find(int x){return x==father[x]?x:father[x]=find(father[x]);}void Union(int x,int y){  father[find(x)]=y;}void dfs(int u){ance[u]=u;int dd=M[u].size();for(int i=0;i<dd;++i){dfs(M[u][i]);Union(u,M[u][i]);ance[find(u)]=u;}vis[u]=1;dd=Q[u].size();for(int i=0;i<dd;i++){if(vis[Q[u][i]]==1){printf("%d\n",ance[find(Q[u][i])]);return;}}}int main(){int t,n,u,v;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<=n;i++){in[i]=0;father[i]=i;vis[i]=false;ance[i]=0;M[i].clear();Q[i].clear();}for(int i=1;i<n;i++){scanf("%d%d",&u,&v);M[u].push_back(v);in[v]++;}scanf("%d%d",&u,&v);Q[u].push_back(v);Q[v].push_back(u);for(int i=1;i<=n;i++)if(in[i]==0){dfs(i);break;}}return 0;}