poj 1330 Nearest Common Ancestors(LCA)

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题目链接：http://poj.org/problem?id=1330

题意：给出一颗树，最后寻问两个点的最近公共祖先。

可以用LCA来解决，没什么问题。

#include<iostream>#include<cstdio>#include<cstring>#include<vector>using namespace std;#define clr(a,b) memset(a,b,sizeof(a))const int M = 10005;int h[M],num[M],nex[M],pos;int father[M],ancestor[M],degree[M];bool flag[M];int s,t;void init(){    clr(degree,0);    clr(flag,false);    clr(h,-1);    pos = 0;    clr(father,-1);}void add(int u,int v){    num[pos] = v;    nex[pos] = h[u];    h[u] = pos++;}void make_set(int x){    father[x] = x;}int find(int x){    if(father[x] == x) return x;    return father[x] = find(father[x]);}void Union(int a,int b){    father[find(b)] = find(a);}void LCA(int u){    ancestor[u] = u;    make_set(u);    for(int i = h[u];i != -1;i = nex[i])    {        LCA(num[i]);        Union(num[i],u);        ancestor[ find(u) ] = u;    }    flag[u] = true;    if(s == u && flag[t])    {        cout<<ancestor[ find(t) ]<<endl;    }    if(t == u && flag[s])    {        cout<<ancestor[ find(s) ]<<endl;    }}int main(){    int n,T;    cin>>T;    while(T--)    {        cin>>n;        init();        int u,v;        for(int i = 0;i < n-1;++i)        {            cin>>u>>v;            add(u,v);            degree[v]++;        }        cin>>s>>t;        for(int i = 1;i <= n;++i)        {            if(!degree[i]) LCA(i);        }    }    return 0;}

参考代码：