lca(倍增)

例题:https://acm.hdu.edu.cn/showproblem.php?pid=2586

点击查看代码

#include <bits/stdc++.h>

using namespace std;

const int N = 40010;

vector<int> v[N];
vector<int> w[N];

int fa[N][31],cost[N][31],dep[N];
int n,m;
int a,b,c;

void dfs(int root, int fno){
    fa[root][0] = fno;
    dep[root] = dep[fa[root][0]] + 1;
    for(int i = 1; i < 31; i ++){
        fa[root][i] = fa[fa[root][i - 1]][i - 1];
        cost[root][i] = cost[fa[root][i - 1]][i - 1] + cost[root][i - 1];
    }
    int sz = v[root].size();
    for(int i = 0; i < sz; i ++){
        if(v[root][i] == fno)continue;
        cost[v[root][i]][0] = w[root][i];
        dfs(v[root][i],root);
    }
}

int lca(int x,int y){
    //令 y 比 x 深
    if(dep[x] > dep[y])swap(x,y);
    int tmp = dep[y] - dep[x];
    int ans = 0;
    for(int j = 0; tmp; j ++, tmp >>= 1){
        if(tmp & 1) ans += cost[y][j],y = fa[y][j];
    }
    if(x == y) return ans;
    for(int j = 30; j >=0 && y != x; j --){
        if(fa[x][j] != fa[y][j]){
            ans += cost[x][j] + cost[y][j];
            x = fa[x][j];
            y = fa[y][j];
        }
    }
    //x和y都在lca(x,y)前一步了
    ans += cost[x][0] + cost[y][0];
    return ans;
}

void solve(){
    memset(fa,0,sizeof(fa));
    memset(cost,0,sizeof(cost));
    memset(dep,0,sizeof(dep));
    
    cin >> n >> m;

    for(int i = 1; i <= n; i ++){
        v[i].clear();
        w[i].clear();
    }

    for(int i = 1; i <= n - 1; i ++){
        cin >> a >> b >> c;
        v[a].push_back(b);
        v[b].push_back(a);
        w[a].push_back(c);
        w[b].push_back(c);
    }

    dfs(1,0);

    for(int i = 0; i < m; i ++){
        cin >> a >> b;
        cout << lca(a,b) << '\n';
    }
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while(t --){
        solve();
    }
    return 0;
}