How far away ? 树上求两点之间最短距离(用LCA)

How far away ?

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
#include <cassert>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-6;
const int mod = 1e9 + 7;

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
inline ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
void exgcd(ll A, ll B, ll& x, ll& y)
{
    if (B) exgcd(B, A % B, y, x), y -= A / B * x; else x = 1, y = 0;
}
inline int qpow(int x, ll n) {
    int r = 1;
    while (n > 0) {
        if (n & 1) r = 1ll * r * x % mod;
        n >>= 1; x = 1ll * x * x % mod;
    }
    return r;
}
inline int inv(int x) {
    return qpow(x, mod - 2);
}
ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
/**********************************************************/
const int N = 4e4 + 5;
int f[N][31],cost[N][31],dep[N];

void dfs(int v, int pa, vector<vector<int>>& g, vector<vector<int>>& w)
{
    f[v][0] = pa;
    dep[v] = dep[pa] + 1;
    for (int i = 1; i < 31; i++)
    {
        f[v][i] = f[f[v][i - 1]][i - 1];//x的2^i祖先是它2^i-1祖先的2^i-1祖先
        cost[v][i] = cost[f[v][i - 1]][i - 1] + cost[v][i - 1];//x到它2^i祖先的距离 = x到它2^i-1祖先的距离 + x的2^i-1祖先到它2^i-1祖先距离
    }
    for (int i = 0; i < g[v].size(); i++)
    {
        int to = g[v][i];
        if (to != pa)
        {
            cost[to][0] = w[v][i];
            dfs(to, v, g, w);
        }
    }
}

int lca(int x, int y)
{
    if (dep[x] > dep[y])
        swap(x, y);
    int tmp = dep[y] - dep[x], ans = 0;
    for (int j = 0; tmp; j++, tmp >>= 1)
        if (tmp & 1)
            ans += cost[y][j], y = f[y][j];
    if (y == x)
        return ans;
    for (int j = 30; j >= 0 && y != x; j--)
    {
        if (f[x][j] != f[y][j])
        {
            ans += cost[x][j] + cost[y][j];
            x = f[x][j];
            y = f[y][j];
        }
    }
    ans += cost[x][0] + cost[y][0];
    return ans;
}

int main() {
    ios::sync_with_stdio(false);cin.tie(nullptr); cout.tie(nullptr);
    int T;
    cin >> T;
    while (T--)
    {
        int n, m;
        cin >> n >> m;
        vector<vector<int>> g(n + 1), w(n + 1);
        rep(i, 1, n - 1)
        {
            int u, v, d;
            cin >> u >> v >> d;
            g[u].push_back(v);
            g[v].push_back(u);
            w[u].push_back(d);
            w[v].push_back(d);
        }
        dfs(1, -1, g, w);
        rep(i, 1, m)
        {
            int u, v;
            cin >> u >> v;
            cout << lca(u, v) << endl;
        }
    }
}
原文地址:https://www.cnblogs.com/dealer/p/13418052.html