hiho16动态lca

这个要好写点，就是dfs这棵树，将每个节点进入 和出去的位置都记录下来，

然后找 所要求得两个点最后出现的位置之间的最小值 所对应的点，就是他们的最近公共祖先。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<vector>
using namespace std;
const int maxn = 111111;

struct edge
{
    int to; int next;
}e[maxn*10];
int len;
int head[maxn];
struct Node
{
    int val; int id;
}vis[maxn],dp[maxn][20];
int pos[maxn];
int cnt;
void add(int from, int to)
{
    e[len].to = to;
    e[len].next = head[from];
    head[from] = len++;
}

void dfs(int x, int val)
{
    vis[cnt].val = val; vis[cnt].id = x; pos[x] = cnt++;
    for (int i = head[x]; i != -1; i = e[i].next){
        int cc = e[i].to;
        dfs(cc, val + 1);
        vis[cnt].val = val; vis[cnt].id = x; pos[x] = cnt++;
    }
}

void init(int k)
{
    for (int i = 0; i < k; i++)
        dp[i][0] = vis[i];
    for (int j = 1; (1 << j) <= k; j++){
        for (int i = 0; i + (1 << j) - 1 < k; i++){
            if (dp[i][j - 1].val < dp[i + (1 << (j - 1))][j - 1].val)
                dp[i][j] = dp[i][j - 1];
            else dp[i][j] = dp[i + (1 << (j - 1))][j - 1];
        }
    }
}

int ask(int l, int r)
{
    int k = 0;
    while ((1 << (k + 1)) < r - l + 1) k++;
    if (dp[l][k].val < dp[r - (1 << k) + 1][k].val) return dp[l][k].id;
    else return dp[r - (1 << k) + 1][k].id;
}
int main()
{
    map<string, int> m;
    map<int, string> m1;
    string a, b;
    int n;
    cin >> n;
    int sum = 1;
    len = 0;
    cnt = 0;
    memset(head, -1, sizeof(head));
    for (int i = 0; i < n; i++){
        cin >> a >> b;
        if (!m.count(a)) m[a] = sum, m1[sum] = a, sum++;
        if (!m.count(b)) m[b] = sum, m1[sum] = b, sum++;
        int  c = m[a]; int d = m[b];
        add(c, d);
    }
    dfs(1, 1);
    init(cnt);
    int q;
    cin >> q;
    while (q--){
        cin >> a >> b; int c = pos[m[a]]; int d = pos[m[b]];
        if (c > d) swap(c, d);
        cout << m1[ask(c, d)] << endl;
    }
    return 0;
}