HDU 6031 Innumerable Ancestors （LCA）

There is a tree having n nodes, labeled from 1 to n. The root of the tree is always 1, and the depth of a node p is the number of nodes on the shortest path between node p and the root.
In computer science, the Lowest Common Ancestor (LCA) of two nodes v and w in a tree is the lowest (i.e. deepest) node that has both v and w as descendants, where we define each node to be a descendant of itself (so if v has a direct connection from w, w is the lowest common ancestor).
You have to answer m queries. Each query gives two non-empty node sets A and B, there might be some nodes in both sets.
You should select one node x from set A, and one node y from set B, x and y can be the same node. Your goal is to maximize the depth of the LCA of x and y.
Please write a program to answer these queries.

题意

给定 N 节点的树，问集合 A 中节点 x 与集合 B 中节点 y 的 LCA(x, y) 的深度的最大值。

解题思路

根据 DFS 序，若两个点的 DFS 序越接近，则两个点的 LCA 的深度越大。

故可将集合 A 与集合 B 中的点分别按照 DFS 序排序。遍历集合 A 中所有元素，对任意一个元素 x，在集合 B 中二分查找与其 DFS 序最接近的点y1, y2 （DFS 序大于等于 x 及 DFS 序小于 x）。利用 LCA 算法 O(logN) 求取 LCA(x, y1), LCA(x, y2) ，并取深度的最大值即可。

代码

#include<bits/stdc++.h>using namespace std;const int N = 100000 + 10;const int maxh = 20; //log2(N);struct Edge {    int to, nxt;} e[N*2];int head[N], cnt;void addedge(int u, int v) {    e[++cnt].nxt = head[u];    e[cnt].to = v;    head[u] = cnt;}int id[N], idx, a[N], b[N], n, m;void genId(int rt, int pre) {    id[rt] = ++idx;    for(int i=head[rt];i;i=e[i].nxt) {        if(e[i].to == pre)  continue;        genId(e[i].to, rt);    }}bool cmp(int a, int b) {    return id[a] < id[b];}int dep[N], anc[N][20];void dfs(int rt) {    static int Stack[N];    int top = 0;    dep[rt] = 1;    for(int i=0;i<maxh;i++)        anc[rt][i] = rt;    Stack[++top] = rt;    while(top) {        int x = Stack[top];        if(x != rt) {            for(int i=1, y;i<maxh;i++)                y = anc[x][i-1],    anc[x][i] = anc[y][i-1];        }        for(int &i=head[x];i;i=e[i].nxt) {            int y = e[i].to;            if(y != anc[x][0]) {                dep[y] = dep[x] + 1;                anc[y][0] = x;                Stack[++top] = y;            }        }        while(top && head[Stack[top]] == 0) top--;    }}void swim(int &x, int H) {    for(int i=0;H;i++) {        if(H & 1)   x = anc[x][i];        H /= 2;    }}int lca(int x, int y) {    int i;    if(dep[x] > dep[y]) swap(x, y);    swim(y, dep[y]-dep[x]);    if(x == y)  return x;    for(;;) {        for(i=0;anc[x][i] != anc[y][i];i++);        if(i == 0)  return anc[x][0];        x = anc[x][i-1];        y = anc[y][i-1];    }    return -1;}void init() {    memset(head, 0, sizeof(head));    cnt = 0;    idx = 0;}int main(){    while(scanf("%d %d", &n, &m)!=EOF)    {        init();        for(int i=1, u, v;i<n;i++)            scanf("%d %d",&u,&v),            addedge(u, v),            addedge(v, u);        genId(1, -1);        dfs(1);        for(int i=1, ka, kb;i<=m;i++)        {            scanf("%d",&ka);            for(int j=0;j<ka;j++)                scanf("%d",&a[j]);            sort(a, a+ka, cmp);            scanf("%d",&kb);            for(int j=0;j<kb;j++)                scanf("%d",&b[j]);            sort(b, b+kb, cmp);            int ans = 0;            for(int j=0, pos;j<ka;j++)            {                pos = lower_bound(b, b+kb, a[j], cmp) - b;                if(pos < kb)                    ans = max(ans, dep[lca(a[j], b[pos])]);                if(--pos >= 0)                    ans = max(ans, dep[lca(a[j], b[pos])]);            }            printf("%d\n", ans);        }    }}