POJ 1330 Nearest Common Ancestors 【用Tarjan求LCA】

题目链接：http://poj.org/problem?id=1330

题意：
有一棵树，n个结点n-1条边，给你这几条边的信息，再给你两个点，求这两个点的LCA（最近公共祖先）
题解：
一道小题，直接用Tarjan就水过去了，注意细节！

代码：

#include <cstdio>#include <vector>#include <cstring>using namespace std;const int size = 3e5+5;vector<int> tree[size];int n, qa, qb, deg[size], vis[size], root;int fath[size];int get(int x) {    if(fath[x] == x) return fath[x];    return fath[x] = get(fath[x]);}void Unionset(int x, int y) {    int fx = get(x), fy = get(y);    if( fx == fy ) return ;    fath[fy] = fx;}void init() {    for ( int i = 0; i <= n; i ++ ) fath[i] = i;}int ans = 0;void Tarjan(int rt) {    for ( int i = 0; i < tree[rt].size(); i ++ ) {        Tarjan(tree[rt][i]);        Unionset(rt, tree[rt][i]);    }    vis[rt] = 1;    if(rt == qa) {        if(vis[qb]) {            ans = get(qb);            return ;        }    }    if(rt == qb) {        if(vis[qa]) {            ans = get(qa);            return ;        }    }}int main() {    // freopen("1330.in","r",stdin);    int tst;    scanf("%d", &tst);    while(tst--){        memset(deg, 0, sizeof(deg));        memset(vis, 0, sizeof(vis));        scanf("%d",&n);        init();        for ( int i = 0; i <= n; i ++ ) tree[i].clear();        for ( int i = 0; i < n-1; i ++ ) {            int u, v;            scanf("%d %d", &u, &v);            tree[u].push_back(v);   // 有向边            deg[v] ++;        }               // 找树根        for ( int i = 1; i <= n; i ++ ) {            if(deg[i] == 0) {                root = i;                break;            }        }        scanf("%d %d", &qa, &qb);        Tarjan(root);        printf("%d\n", ans);    }    return 0;}