LCA(树上倍增)
作者:互联网
https://www.luogu.com.cn/problem/P3379
- 链式前向星存边
- fa[i][j] 代表从i结点向上找 2^i 代的父亲,(i=0代表真父亲)
- dfs从根结点开始
fa[now][i] = fa[fa[now][i - 1]][i - 1];
代表当前结点的第2^i代父节点是当前结点2^(i-1)父节点的2^(i-1)代父节点,然后再对其连接到的非父结点dfs - depth记录深度,用于判断循环的次数即log2(depth【x】)
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000001
long long idx;
long long head[MAX];
long long depth[MAX];
long long fa[MAX][21];
long long n, m, s;
long long lg[MAX];
struct node
{
long long to, nex;
} edges[MAX];
void add(long long u, long long v)
{
edges[++idx].to = v;
edges[idx].nex = head[u];
head[u] = idx;
}
void input()
{
cin >> n >> m >> s;
for (long long i = 0; i < n - 1; i++)
{
long long a, b;
scanf("%lld%lld", &a, &b);
add(a, b);
add(b, a);
}
}
void dfs(long long now, long long faer)
{
fa[now][0] = faer;
depth[now] = depth[faer] + 1;
for (long long i = 1; i <= log2(depth[now]); i++)
{
fa[now][i] = fa[fa[now][i - 1]][i - 1];
}
for (long long i = head[now]; i; i = edges[i].nex)
{
if (edges[i].to != faer)
{
dfs(edges[i].to, now);
}
}
}
long long LCA(long long a, long long b)
{
if (depth[a] < depth[b])
{
swap(a, b);
}
while (depth[a] > depth[b])
{
a = fa[a][lg[depth[a] - depth[b]] - 1];
}
if (a == b)
{
return a;
}
for (long long i = lg[depth[a]] - 1; i >= 0; i--)
{
if (fa[a][i] != fa[b][i])
{
a = fa[a][i];
b = fa[b][i];
}
}
return fa[a][0];
}
void init_log2()
{
for (long long i = 1; i <= n; ++i)
lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
}
void main_work()
{
for (long long i = 0; i < m; i++)
{
long long a, b;
scanf("%lld%lld", &a, &b);
printf("%lld\n", LCA(a, b));
}
}
void print_log2()
{
for (long long i = 0; i <= n; i++)
{
printf("%lld ", lg[i]);
}
}
int main()
{
input();
init_log2();
// print_log2();
dfs(s, 0);
main_work();
}
标签:now,倍增,MAX,结点,long,fa,depth,LCA,树上 来源: https://www.cnblogs.com/Wang-Xianyi/p/16538014.html