POJ 1330 Nearest Common Ancestors【LCA】

题目链接：

http://poj.org/problem?id=1330

题意：

裸的LCA

代码：

倍增：

#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 10005;
const int maxm = 20;
struct Edge{int to, next;};
Edge edge[maxn * 2];
int head[maxn], tot;
int pa[maxm][maxn];
int dept[maxn], in[maxn];
int root, V;
void add_edge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void initedge()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}
//DFS预处理所有结点的深度和父节点
void dfs(int v, int p, int d)
{
    pa[0][v] = p;
    dept[v] = d;
    for(int i = head[v]; i != -1; i = edge[i].next){
        int u = edge[i].to;
        if(u == p) continue;
        dfs(u, v, d + 1);
    }
}
void init()
{
    dfs(root, -1, 0);
    //预处理祖先，向上走2^i所到的结点
    for(int i = 0; i < maxm - 1; i++){
        for(int j = 1; j <= V; j++){
            if(pa[i][j] < 0) pa[i + 1][j] = -1;
            else pa[i + 1][j] = pa[i][pa[i][j]];
        }
    }
}
int lca(int u, int v)
{
    //让u和v 向上走到同一高度
    if(dept[u] > dept[v]) swap(u, v);
    for(int i = 0; i < maxm; i++){
        if((dept[v] - dept[u]) >>i &1)
            v = pa[i][v];
    }
    if(u == v) return u;

    //二分搜索计算lca
    for(int i = maxm - 1; i >= 0; i--){
        if(pa[i][u] != pa[i][v]){
            u = pa[i][u];
            v = pa[i][v];
        }
    }
    return pa[0][u];
}
int main(void)
{
    int T;cin>>T;
    int u, v;
    while(T--){
        cin>>V;
        initedge();
        memset(in, 0, sizeof(in));
        for(int i = 1; i < V; i++){
            cin>>u>>v;
            add_edge(u, v);
            add_edge(v, u);
            in[v] = 1;
        }
        for(root = 1; root <= V; root++){
            if(!in[root]) break;
        }
        init();
        cin>>u>>v;
        cout<<lca(u, v)<<endl;;
    }
    return 0;

RMQ：

（照着bin神的模板敲的

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn = 10005;
const int maxm = 20;
struct Edge{int to, next;};
Edge edge[maxn * 2];
int head[maxn], tot;
int pa[maxm][maxn];
int dept[maxn], in[maxn];
int root, V, cnt;
int vs[maxn * 2], id[maxn];
int rmq[2 * maxn];
void add_edge(int u, int v)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
void initedge()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}
struct ST
{
    int lg[2 * maxn];
    int dp[2 * maxn][20];//最小值对应下标
    void init(int n){
        lg[0] = -1;
        for(int i = 1; i <= n; i++){
            lg[i] = ((i & (i - 1)) == 0)?lg[i - 1] + 1:lg[i - 1];
            dp[i][0] = i;
        }
        for(int j = 1; j <= lg[n]; j++){
            for(int i = 1; i + (1<<j) - 1 <= n; i++){
                if(rmq[dp[i][j - 1]] < rmq[dp[i + (1<<(j - 1))][j - 1]])
                    dp[i][j] = dp[i][j - 1];
                else dp[i][j] = dp[i + (1<<(j - 1))][j - 1];
            }
        }
    }
    int query(int a, int b){
        if(a > b) swap(a, b);
        int k = lg[b - a + 1] ;
       // cout<<b<<' '<<a<<endl;
        if(rmq[dp[a][k]] <= rmq[dp[b - (1<<k) + 1][k]])
            return dp[a][k];
        else return dp[b - (1<<k) + 1][k];
    }
};
ST st;
//vs dfs的访问顺序
void dfs(int u, int pre, int dept)
{
    vs[++cnt] = u;
    rmq[cnt] = dept;
    id[u] = cnt;
    int fro = head[u];
    while(fro != -1){
        dfs(edge[fro].to, u, dept + 1);
        fro = edge[fro].next;
        vs[++cnt] = u;
        rmq[cnt] = dept;
    }
}
void init()
{
    cnt = 0;
    dfs(root, root, 0);
    st.init(2 * V - 1);
}
int lca(int u, int v)
{
    return vs[st.query(id[u], id[v])];
}
int main(void)
{
    int T;cin>>T;
    int u, v;
    while(T--){
        cin>>V;
        initedge();
        memset(in, 0, sizeof(in));
        for(int i = 1; i < V; i++){
            cin>>u>>v;
            add_edge(u, v);
            in[v] = 1;
        }
        for(root = 1; root <= V; root++){
            if(!in[root]) break;
        }
        init();
        cin>>u>>v;
        cout<<lca(u, v)<<endl;;
    }
    return 0;
}