Tarjan求Lca

https://segmentfault.com/a/1190000015145319?utm_source=index-hottest

//参考题目洛谷P3379【模板】最近公共祖先

#include <vector>
#include <cstdio>
using namespace std;
inline int read()
{
    int x = 0, f = 0;
    char c = getchar();
    for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = 1;
    for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ '0');
    return f ? -x : x;
}

const int N = 5e5 + 7;

int n, m, u, v, s;
int tot = 0, st[N], to[N << 1], nx[N << 1], fa[N], ans[N], vis[N];
struct note 
{ 
int node, id; 
}; //询问以结构体形式保存
vector<note> ques[N];

inline void add(int u, int v) 
{ 
to[++tot] = v, nx[tot] = st[u], st[u] = tot; 
}
inline int getfa(int x) 
{ 
return fa[x] == x ? x : fa[x] = getfa(fa[x]); 
} //并查集的getfa操作，路径压缩

void dfs(int u, int from)
{
    for (int i = st[u]; i; i = nx[i]) 
	if (to[i] != from) 
	dfs(to[i], u), fa[to[i]] = u; 
	//将u的儿子合并到u,这一句话很重要to[i]这个点的父亲最开始为其自身，现在要改为u了 
    int len = ques[u].size(); 
	//处理与u有关的询问
    for (int i = 0; i < len; i++) 
	if (vis[ques[u][i].node]) 
	//目前在u点上，如果针对u的询问中，另一个点v是那种已完全访问过的的话 
	ans[ques[u][i].id] = getfa(ques[u][i].node); 
	//对应的v已经访问并回溯时，LCA(u,v)就是v的fa里深度最小的一个也就是getfa(v)
    vis[u] = 1; //访问完毕回溯
}

int main()
{
    n = read(), m = read(), s = read();
    for (int i = 1; i < n; i++) 
	u = read(), v = read(), add(u, v), add(v, u);
    for (int i = 1; i <= m; i++) 
	    u = read(), v = read(), 
	    ques[u].push_back((note){v, i}), 
	    ques[v].push_back((note){u, i});
	 //记下询问编号便于输出
    for (int i = 1; i <= n; i++) 
    	fa[i] = i;
    dfs(s, 0);
    for (int i = 1; i <= m; i++) 
	printf("%d\n", ans[i]); //输出答案
    return 0;
}