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 25
using 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;
}
posted @ 2017-08-11 19:16  pengwill  阅读(89)  评论(0编辑  收藏  举报