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;

}