poj 1330 Nearest Common Ancestors（LCA）_nearest common ancestors acm-优快云博客

本文链接：https://blog.youkuaiyun.com/keshuai19940722/article/details/49799561

本文介绍了一种求解最近公共祖先(LCA)问题的有效算法。通过预处理构建稀疏表格，实现快速查询。代码中利用深度优先搜索计算节点深度，并通过队列进行层次遍历初始化稀疏表。

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代码

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;
const int maxn = 1e4 + 5;
const int inf = 0x3f3f3f3f;
const int BIT = 20;

int dep[maxn], f[maxn][BIT+5];
vector<int> G[maxn];

void ST (int n) {
    for (int k = 1; k <= BIT; k++) {
        for (int i = 1; i <= n; i++) 
            f[i][k] = f[f[i][k-1]][k-1];
    }

    int s;
    for (int i = 1; i <= n; i++) if (f[i][0] == 0) {
        s = i; break;
    }
    memset(dep, inf, sizeof(dep));
    dep[s] = 0;

    queue<int> que;
    que.push(s);
    while (!que.empty()) {
        int u = que.front();
        que.pop();

        for (int i = 0; i < G[u].size(); i++) {
            int v = G[u][i];
            if (dep[v] > dep[u] + 1) {
                dep[v] = dep[u] + 1;
                que.push(v);
            }
        }
    }
}

int LCA(int u, int v) {
    if (dep[u] > dep[v]) swap(u, v);

    int mv = dep[v] - dep[u];
    for (int i = 0; i <= BIT && mv; mv >>= 1, i++)
        if (mv&1) v = f[v][i];

    if (u == v) return u;

    for (int i = BIT; i >= 0; i--) {
        if (f[u][i] == f[v][i]) continue;
        u = f[u][i]; v = f[v][i];
    }
    return f[u][0];
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        memset(f, 0, sizeof(f));

        int n, u, v;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) G[i].clear();
        for (int i = 1; i < n; i++) {
            scanf("%d%d", &u, &v);
            f[v][0] = u;
            G[u].push_back(v);
        }
        ST(n);
        scanf("%d%d", &u, &v);
        printf("%d\n", LCA(u, v));
    }
    return 0;
}