• 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;
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/13418052.html
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