POJ.1330 Nearest Common Ancestors (LCA 倍增)

POJ.1330 Nearest Common Ancestors (LCA 倍增)

题意分析

给出一棵树，树上有n个点(n-1)条边，n-1个父子的边的关系a-b。接下来给出xy,求出xy的lca节点编号。

LCA裸题，用倍增思想。

代码总览

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#define nmax 80520#define demen 25using namespace std;int fa[nmax][demen],head[nmax],dep[nmax];int n,m,tot = 0;struct node{    int to;    int next;    int w;}edge[nmax];void add(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void dfs(int rt,int f){    fa[rt][0] = f;    for(int i = 1;i<=20;++i){        fa[rt][i] = fa[fa[rt][i-1]][i-1];    }    for(int i = head[rt];i!=-1;i = edge[i].next){        int nxt = edge[i].to;        if(nxt != f){            dep[nxt] = dep[rt] + 1;            dfs(nxt,rt);        }    }}int lca(int x, int y){    int X = x,Y=y;    if(dep[x] < dep[y]) swap(x,y);    int dis = dep[x] - dep[y];    for(int i = 20;i>=0;--i){        if((1<<i) & dis)            x = fa[x][i];    }    if(x == y) return(x);    for(int i = 20;i>=0;--i){        if(fa[x][i] != fa[y][i]){            x = fa[x][i],y = fa[y][i];        }    }    return(fa[x][0]);}void init(){    memset(fa,0,sizeof fa);    memset(head,-1,sizeof head);    memset(dep,0,sizeof dep);    tot = 0;}int main(){    int t;    scanf("%d",&t);    while(t--){        init();        int n,u,v;        scanf("%d",&n);        int root = 0;        for(int i = 0;i<n-1;++i){            scanf("%d %d",&u,&v);            if(root == 0) root = u;            add(u,v);            add(v,u);        }        dep[root] = 1;        dfs(root,0);        scanf("%d %d",&u,&v);        printf("%d\n",lca(u,v));    }    return 0;}