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  • CF425 D. Misha, Grisha and Underground

      我天,这竟然是道最近公共祖先的板子题。代码里利用了我的LCA模板,前向星存图。

     Misha, Grisha and Underground

    #define _CRT_SECURE_NO_WARNINGS
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <iostream>
    using namespace std;
    const int N = 1e5 + 7, H = 20;
    int n, tot; //树的根节点
    int head_copy[N], head[N]; //存的边的信息,head_copy[i]表示第i个节点的头指针
    int to[N << 1]; //第i条边指向的节点
    int nxt[N << 1]; //第i条边的下一个指针
    int anc[N][H]; //对于每一个节点v,记录anc[v][k],表示它向上走pow(2,k)步之后到达的节点
    int Stack[N], dep[N];
    void dfs(int root) {
        int top = 0;
        dep[root] = 1;
        for (int i = 0; i < H; ++i)
            anc[root][i] = root;
        Stack[++top] = root; //先求出anc[v][0]
        memcpy(head_copy, head, sizeof head);//head为原始的第i个节点的头指针
        while (top) {
            int x = Stack[top];
            if (x != root) {
                for (int i = 1; i < H; ++i) { //再求出其他anc[v][k]
                    int y = anc[x][i - 1];
                    anc[x][i] = anc[y][i - 1];
                }
            }
            for (int &i = head_copy[x]; ~i; i = nxt[i]) { //这里i为引用,会修改head_copy
                int y = to[i];
                if (y != anc[x][0]) {
                    dep[y] = dep[x] + 1;
                    anc[y][0] = x;
                    Stack[++top] = y;
                }
            }
            while (top && head_copy[Stack[top]] == -1) //==-1,这个-1和初始head有关
                top--;
        }
    }
    inline void swim(int &x, int k) { //从节点x向上移动k步,并将x赋为新走到的节点
        for (int i = 0; k > 0; ++i) {
            if (k & 1)
                x = anc[x][i];
            k /= 2;
        }
    }
    int lca(int x, int y) { //寻找x, y的LCA。
        int k;
        if (dep[x] > dep[y]) swap(x, y);
        swim(y, dep[y] - dep[x]); //首先利用swim将x,y调整到同一高度
        if (x == y) return x; //若x和y重合,就是我们要找的LCA
        while (true) { //否则,不断第寻找一个最小的k,使得anc[x][k] = anc[y][k]
            for (k = 0; anc[x][k] != anc[y][k]; ++k);
            if (k == 0)
                return anc[x][0];
            x = anc[x][k - 1];//新的x,y和原来的x,y有相同的LCA
            y = anc[y][k - 1];
        }
        return -1;
    }
    void init() {
        tot = 0;
        memset(head, -1, sizeof head);
    }
    void add(int u, int v) {
        to[tot] = v;
        nxt[tot] = head[u];
        head[u] = tot++;
    }
    int dist(int u, int v) {
        int x = lca(u, v);
        int res = dep[u] + dep[v] - 2 * dep[x];
        return res;
    }
    int main()
    {
        ios::sync_with_stdio(false);
        int q, pi;
        while (cin >> n >> q) {
            init();
            for (int i = 2; i <= n; ++i) {
                cin >> pi;
                add(pi, i);
                add(i, pi);
            }
            dfs(1);
            int a, b, c, m1, m2, m3, m;
            while (q--) {
                cin >> a >> b >> c;
                m1 = lca(a, b), m2 = lca(a, c), m3 = lca(b, c);
                if (m1 == m2) m = m3;
                else if (m1 == m3) m = m2;
                else m = m1;
                cout << 1 + max(dist(m, a), max(dist(m, b), dist(m, c))) << endl;
            }
        }
    
        return 0;
    }

    by myorange

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  • 原文地址:https://www.cnblogs.com/qinwenjie/p/7267644.html
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