POJ 1330 Nearest Common Ancestors【LCA】

题目链接：

题意：

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：

#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
; 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;
}