hdu 2586 How far away ？ 倍增求LCA

倍增求LCA

LCA函数返回(u,v)两点的最近公共祖先

#include <bits/stdc++.h>
using namespace std;

const int N = 40010*2;
struct node {
    int v,val,next;
    node(){}
    node(int vv,int va,int nn):v(vv),val(va),next(nn){}
}E[N];
int n,m;
int tot,head[N],dis[N],f[N][20],dep[N];

void init() {
    tot = 0;
    memset(head,0,sizeof(head));
    memset(dis,0,sizeof(dis));
    memset(f,0,sizeof(f));
    memset(dep,0,sizeof(dep));
}

void add(int u,int v,int val) {
    E[++tot].next = head[u];
    E[tot].v = v;
    E[tot].val = val;
    head[u] = tot;
}

void addEdge(int u,int v,int val) {
    add(u,v,val);
    add(v,u,val);
}

void dfs(int x,int fa) {
    f[x][0] = fa;
    for(int i=1;f[x][i-1];i++)
        f[x][i] = f[f[x][i-1]][i-1];
    for(int i=head[x]; i; i=E[i].next) {
        int v = E[i].v;
        if(v != fa) {
            dis[v] = dis[x] + E[i].val;
            dep[v] = dep[x] + 1;
            dfs(v,x);
        }
    }
}

int lca(int u,int v) {
    if(dep[u] < dep[v]) swap(u,v);
    //int ans = dep[u] - dep[v];
    for(int i=19;i>=0;i--)
        if(dep[u]-(1<<i)>=dep[v])
            u = f[u][i];
    for(int i=19;i>=0;i--) {
        if(f[u][i]!=f[v][i])
            u=f[u][i],v=f[v][i];//ans+=(2<<i);
    }
    //if(u!=v) ans+=2;
    //return ans;
    if(u!=v) return f[u][0];
    else return u;
}

int main () {
    //freopen("in.txt","r",stdin);
    int T; scanf("%d",&T);
    while (T--) {
        init();
        scanf("%d %d",&n, &m);
        for(int i=1;i<=n-1;i++) {
            int u,v,val;
            scanf("%d %d %d",&u,&v,&val);
            addEdge(u,v,val);
        }
        dfs(1,0);
        // cout <<"yes" <<endl;
        while(m--) {
            int u,v; scanf("%d %d",&u,&v);
            int  t = lca(u,v);
            printf("%d\n",dis[u]+dis[v]-2*dis[t]);
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/Draymonder/p/9617303.html