DFS + ST在线算法
参考题目链接：
POJ 1330 Nearest Common Ancestors
const int MAXN = 10010;
int rmq[2 * MAXN]; // rmq数组,就是欧拉序列对应的深度序列
struct ST
{
int mm[2 * MAXN];
int dp[2 * MAXN][20]; // 最小值对应的下标
void init(int n)
{
mm[0] = -1;
for (int i = 1; i <= n; i++)
{
mm[i] = ((i & (i - 1)) == 0) ? mm[i - 1] + 1 : mm[i - 1];
dp[i][0] = i;
}
for (int j = 1; j <= mm[n]; j++)
{
for (int i = 1; i + (1 << j) - 1 <= n; i++)
{
dp[i][j] = rmq[dp[i][j - 1]] < rmq[dp[i + (1 << (j - 1))][j - 1]] ? dp[i][j - 1] : dp[i + (1 << (j - 1))][j - 1];
}
}
}
int query(int a,int b) // 查询[a,b]之间最小值的下标
{
if (a > b)
{
swap(a, b);
}
int k = mm[b - a + 1];
return rmq[dp[a][k]] <= rmq[dp[b - (1 << k) + 1][k]] ? dp[a][k] : dp[b - (1 << k) + 1][k];
}
};
// 边的结构体定义
struct Edge
{
int to, next;
};
Edge edge[MAXN * 2];
int tot, head[MAXN];
int F[MAXN * 2]; // 欧拉序列,就是dfs遍历的顺序,长度为2*n-1,下标从1开始
int P[MAXN]; // P[i]表示点i在F中第一次出现的位置
int cnt;
ST st;
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v) // 加边,无向边需要加两次
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void dfs(int u, int pre, int dep)
{
F[++cnt] = u;
rmq[cnt] = dep;
P[u] = cnt;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (v == pre)
{
continue;
}
dfs(v, u, dep + 1);
F[++cnt] = u;
rmq[cnt] = dep;
}
}
void LCA_init(int root, int node_num) // 查询LCA前的初始化
{
cnt = 0;
dfs(root, root, 0);
st.init(2 * node_num - 1);
}
int query_lca(int u, int v) // 查询u,v的lca编号
{
return F[st.query(P[u], P[v])];
}
bool flag[MAXN];
int main()
{
int T;
int N;
int u, v;
scanf("%d", &T);
while(T--)
{
scanf("%d", &N);
init();
memset(flag, false, sizeof(flag));
for (int i = 1; i < N; i++)
{
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
flag[v] = true;
}
int root;
for (int i = 1; i <= N; i++)
{
if (!flag[i])
{
root = i;
break;
}
}
LCA_init(root, N);
scanf("%d%d", &u, &v);
printf("%d\n", query_lca(u, v));
}
return 0;
}Tarjan离线算法
参考题目链接：
POJ 1470 Closest Common Ancestors
/* * 给出一颗有向树，Q个查询 * 输出查询结果中每个点出现次数 * 复杂度O(n + Q); */
const int MAXN = 1010;
const int MAXQ = 500010; // 查询数的最大值
// 并查集部分
int F[MAXN]; // 需要初始化为-1
int find(int x)
{
if (F[x] == -1)
{
return x;
}
return F[x] = find(F[x]);
}
void bing(int u, int v)
{
int t1 = find(u);
int t2 = find(v);
if (t1 != t2)
{
F[t1] = t2;
}
}
bool vis[MAXN]; // 访问标记
int ancestor[MAXN]; // 祖先
struct Edge
{
int to, next;
} edge[MAXN * 2];
int head[MAXN],tot;
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
struct Query
{
int q, next;
int index; // 查询编号
} query[MAXQ * 2];
int answer[MAXQ]; // 存储最后的查询结果,下标0~Q-1
int h[MAXQ];
int tt;
int Q;
void add_query(int u, int v, int index)
{
query[tt].q = v;
query[tt].next = h[u];
query[tt].index = index;
h[u] = tt++;
query[tt].q = u;
query[tt].next = h[v];
query[tt].index = index;
h[v] = tt++;
}
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
tt = 0;
memset(h, -1, sizeof(h));
memset(vis, false, sizeof(vis));
memset(F, -1, sizeof(F));
memset(ancestor, 0, sizeof(ancestor));
}
void LCA(int u)
{
ancestor[u] = u;
vis[u] = true;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (vis[v])
{
continue;
}
LCA(v);
bing(u, v);
ancestor[find(u)] = u;
}
for (int i = h[u]; i != -1; i = query[i].next)
{
int v = query[i].q;
if (vis[v])
{
answer[query[i].index] = ancestor[find(v)];
}
}
}
bool flag[MAXN];
int Count_num[MAXN];
int main()
{
int n;
int u, v, k;
while (scanf("%d", &n) == 1)
{
init();
memset(flag, false, sizeof(flag));
for (int i = 1; i <= n; i++)
{
scanf("%d:(%d)", &u, &k);
while (k--)
{
scanf("%d", &v);
flag[v] = true;
addedge(u,v);
addedge(v,u);
}
}
scanf("%d", &Q);
for (int i = 0; i < Q; i++)
{
char ch;
cin >> ch;
scanf("%d %d)", &u, &v);
add_query(u, v, i);
}
int root;
for (int i = 1; i <= n; i++)
{
if (!flag[i])
{
root = i;
break;
}
}
LCA(root);
memset(Count_num, 0, sizeof(Count_num));
for (int i = 0; i < Q; i++)
{
Count_num[answer[i]]++;
}
for (int i = 1; i <= n; i++)
{
if (Count_num[i] > 0)
{
printf("%d:%d\n", i, Count_num[i]);
}
}
}
return 0;
}倍增法
POJ 1330 Nearest Common Ancestors
/* * LCA在线算法(倍增法) */
const int MAXN = 10010;
const int DEG = 20;
struct Edge
{
int to, next;
} edge[MAXN * 2];
int head[MAXN], tot;
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
}
int fa[MAXN][DEG]; // fa[i][j]表示结点i的第2^j个祖先
int deg[MAXN]; // 深度数组
void BFS(int root)
{
queue<int>que;
deg[root] = 0;
fa[root][0] = root;
que.push(root);
while (!que.empty())
{
int tmp = que.front();
que.pop();
for (int i = 1; i < DEG; i++)
{
fa[tmp][i] = fa[fa[tmp][i - 1]][i - 1];
}
for (int i = head[tmp]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (v == fa[tmp][0])
{
continue;
}
deg[v] = deg[tmp] + 1;
fa[v][0] = tmp;
que.push(v);
}
}
}
int LCA(int u, int v)
{
if (deg[u] > deg[v])
{
swap(u, v);
}
int hu = deg[u], hv = deg[v];
int tu = u, tv = v;
for (int det = hv-hu, i = 0; det ; det >>= 1, i++)
{
if (det & 1)
{
tv = fa[tv][i];
}
}
if (tu == tv)
{
return tu;
}
for (int i = DEG - 1; i >= 0; i--)
{
if (fa[tu][i] == fa[tv][i])
{
continue;
}
tu = fa[tu][i];
tv = fa[tv][i];
}
return fa[tu][0];
}
bool flag[MAXN];
int main()
{
int T;
int n;
int u, v;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
init();
memset(flag, false, sizeof(flag));
for (int i = 1; i < n; i++)
{
scanf("%d%d", &u, &v);
addedge(u, v);
addedge(v, u);
flag[v] = true;
}
int root;
for (int i = 1; i <= n; i++)
{
if (!flag[i])
{
root = i;
break;
}
}
BFS(root);
scanf("%d%d", &u, &v);
printf("%d\n", LCA(u, v));
}
return 0;
}
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