poj1330 Nearest Common Ancestors

题意：

LCA裸题。

思路：

1. 朴素

2. 基于二分

3. 基于RMQ

实现：

1.

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;

vector<int> G[10005];
int t, n, x, y;
int in[10005], parent[10005], depth[10005];

void dfs(int v, int p, int d)
{
    parent[v] = p;
    depth[v] = d;
    for (int i = 0; i < G[v].size(); i++)
    {
        if (G[v][i] != p)
            dfs(G[v][i], v, d + 1);
    }
}

void init(int root)
{
    dfs(root, -1, 0);
}

int lca(int root, int x, int y)
{
    while (depth[x] > depth[y])
    {
        x = parent[x];
    }
    while (depth[y] > depth[x])
    {
        y = parent[y];
    }
    while (x != y)
    {
        x = parent[x];
        y = parent[y];
    }
    return x;
}

int main()
{
    cin >> t;
    while (t--)
    {
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            G[i].clear();
        }
        memset(depth, 0, sizeof(depth));
        memset(in, 0, sizeof(depth));
        memset(parent, 0, sizeof(parent));
        for (int i = 0; i < n - 1; i++)
        {
            scanf("%d %d", &x, &y);
            G[x].push_back(y);
            in[y]++;
        }
        int root = 0;
        for (int i = 1; i <= n; i++)
        {
            if (!in[i])
            {
                root = i;
                break;
            }
        }
        cin >> x >> y;
        init(root);
        cout << lca(root, x, y) << endl;
    }
    return 0;
}

2.

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;

const int MAX_LOG_N = 14;

vector<int> G[10005];
int t, n, x, y;
int in[10005], parent[MAX_LOG_N][10005], depth[10005];

void dfs(int v, int p, int d)
{
    parent[0][v] = p;
    depth[v] = d;
    for (int i = 0; i < G[v].size(); i++)
    {
        if (G[v][i] != p)
            dfs(G[v][i], v, d + 1);
    }
}

void init(int root)
{
    dfs(root, -1, 0);
    for (int k = 0; k < MAX_LOG_N - 1; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            if (parent[k][i] < 0)
            {
                parent[k + 1][i] = -1;
            }
            else
            {
                parent[k + 1][i] = parent[k][parent[k][i]];
            }
        }
    }
}

int lca(int root, int x, int y)
{
    if (depth[x] > depth[y])
        swap(x, y);
    for (int k = 0; k < MAX_LOG_N; k++)
    {
        if ((depth[y] - depth[x]) >> k & 1)
        {
            y = parent[k][y];
        }
    }
    if (x == y)
    {
        return x;
    }
    for (int k = MAX_LOG_N - 1; k >= 0; k--)
    {
        if (parent[k][x] != parent[k][y])
        {
            x = parent[k][x];
            y = parent[k][y];
        }
    }
    return parent[0][y];
}

int main()
{
    cin >> t;
    while (t--)
    {
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            G[i].clear();
        }
        memset(depth, 0, sizeof(depth));
        memset(in, 0, sizeof(depth));
        memset(parent, 0, sizeof(parent));
        for (int i = 0; i < n - 1; i++)
        {
            scanf("%d %d", &x, &y);
            G[x].push_back(y);
            in[y]++;
        }
        int root = 0;
        for (int i = 1; i <= n; i++)
        {
            if (!in[i])
            {
                root = i;
                break;
            }
        }
        cin >> x >> y;
        init(root);
        cout << lca(root, x, y) << endl;
    }
    return 0;
}

3.

#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

const int MAXN = 10005;
vector<int> G[MAXN];
int t, n, x, y;
int in[MAXN], depth[2 * MAXN - 1], id[MAXN], vs[2 * MAXN - 1];

const int MAXM = 32768;
const int INF = 0x3f3f3f3f;
struct node
{
    int index, d;
};
node data[2 * MAXM - 1];

void update(int k, int a)
{
    int tmp = k;
    k += n - 1;
    data[k].index = tmp;
    data[k].d = a;
    while (k)
    {
        k = (k - 1) / 2;
        if (data[2 * k + 1].d < data[2 * k + 2].d)
        {
            data[k].d = data[2 * k + 1].d;
            data[k].index = data[2 * k + 1].index;
        }
        else
        {
            data[k].d = data[2 * k + 2].d;
            data[k].index = data[2 * k + 2].index;
        }
    }
}

void rmq_init(int * depth, int k)
{
    n = 1;
    while (n < k) n <<= 1;
    for (int i = 0; i < 2 * n; i++)
    {
        data[i].d = INF;
    }
    for (int i = 0; i < k; i++)
    {
        update(i, *(depth + i));
    }
}

node query(int a, int b, int k, int l, int r)
{
    if (b <= l || a >= r)
    {
        node res;
        res.index = -1;
        res.d = INF;
        return res;
    }
    if (l >= a && r <= b)
        return data[k];
    node x = query(a, b, 2 * k + 1, l, (l + r) / 2);
    node y = query(a, b, 2 * k + 2, (l + r) / 2, r);
    if (x.d < y.d)
        return x;
    return y;
}

void dfs(int v, int p, int d, int & k)
{
    id[v] = k;
    vs[k] = v;
    depth[k++] = d;
    for (int i = 0; i < G[v].size(); i++)
    {
        if (G[v][i] != p)
        {
            dfs(G[v][i], v, d + 1, k);
            vs[k] = v;
            depth[k++] = d;
        }
    }
}

int init(int root)
{
    int k = 0;
    dfs(root, -1, 0, k);
    rmq_init(depth, k);
    return k;
}

int lca(int root, int x, int y)
{
    node res = query(min(id[x], id[y]), max(id[x], id[y]) + 1, 0, 0, n);
    return vs[res.index];
}

int main()
{
    cin >> t;
    while (t--)
    {
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
            G[i].clear();
        }
        memset(depth, 0x3f, sizeof(depth));
        memset(in, 0, sizeof(depth));
        memset(id, 0, sizeof(id));
        memset(vs, 0, sizeof(vs));
        for (int i = 0; i < n - 1; i++)
        {
            scanf("%d %d", &x, &y);
            G[x].push_back(y);
            in[y]++;
        }
        int root = 0;
        for (int i = 1; i <= n; i++)
        {
            if (!in[i])
            {
                root = i;
                break;
            }
        }
        int res = init(root);
        cin >> x >> y;
        cout << lca(root, x, y) << endl;
    }
    return 0;
}

总结：

还可以使用tarjan算法。