倍增法求LCA

倍增法求LCA

可以利用LCA快速的求出树上两点间的最短路

时间复杂度O(nlogn)

#include<iostream>
#include<string.h>
#include<algorithm>
#include<vector>
#include<map>
#include<bitset>
#include<set>
#include<math.h>
#include<string>
#if !defined(_WIN32)
#include<bits/stdc++.h>
#endif // !defined(_WIN32)
#define ll long long
#define dd double
using namespace std;
struct edge
{
    int to;
    int weight;
    int next;
}e[500005];
int head[100086];
int p[100086][25];
int deep[100086];
ll dis[100086];
int n, m;
int tot;
void add(int x, int y, int z)
{
    tot++;
    e[tot].to = y;
    e[tot].weight = z;
    e[tot].next = head[x];
    head[x] = tot;
}
void dfs(int x, int f)
{
    p[x][0] = f;
    deep[x] = deep[f] + 1;
    for (int i = 1; (1 << i) <= deep[x]; i++)
        p[x][i] = p[p[x][i - 1]][i - 1];
    for (int i = head[x]; i; i = e[i].next)
    {
        int to = e[i].to;
        if (to != f)
        {
            dis[to] = dis[x] + e[i].weight;
            dfs(to, x);
        }
    }
}
int lca(int x, int y)
{
    if (deep[x] < deep[y])
        swap(x, y);
    while (deep[x] > deep[y])
        x = p[x][(int)log2(deep[x] - deep[y])];
    if (x == y)
        return y;
    for (int k = (int)log2(deep[x]); k >= 0; k--)
        if (p[x][k] != p[y][k])
        {
            x = p[x][k];
            y = p[y][k];
        }
    return p[x][0];
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i < n; i++)
    {
        int x, y, z;
        cin >> x >> y >> z;
        add(x, y, z);
        add(y, x, z);
    }
    dfs(1, 0);
    for (int i = 1; i <= m; i++)
    {
        int x, y;
        cin >> x >> y;
        int c = lca(x, y);
        cout << dis[x] + dis[y] - 2 * dis[c] << endl;
    }
    return 0;
}