POJ 1330 Nearest Common Ancestors(LCA)
Description:
A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:
In the figure, each node is labeled with an integer from {1, 2,…,16}. Node 8 is the root of the tree. Node x is an ancestor of node y if node x is in the path between the root and node y. For example, node 4 is an ancestor of node 16. Node 10 is also an ancestor of node 16. As a matter of fact, nodes 8, 4, 10, and 16 are the ancestors of node 16. Remember that a node is an ancestor of itself. Nodes 8, 4, 6, and 7 are the ancestors of node 7. A node x is called a common ancestor of two different nodes y and z if node x is an ancestor of node y and an ancestor of node z. Thus, nodes 8 and 4 are the common ancestors of nodes 16 and 7. A node x is called the nearest common ancestor of nodes y and z if x is a common ancestor of y and z and nearest to y and z among their common ancestors. Hence, the nearest common ancestor of nodes 16 and 7 is node 4. Node 4 is nearer to nodes 16 and 7 than node 8 is.
For other examples, the nearest common ancestor of nodes 2 and 3 is node 10, the nearest common ancestor of nodes 6 and 13 is node 8, and the nearest common ancestor of nodes 4 and 12 is node 4. In the last example, if y is an ancestor of z, then the nearest common ancestor of y and z is y.
Write a program that finds the nearest common ancestor of two distinct nodes in a tree.
Input:
The input consists of T test cases. The number of test cases (T) is given in the first line of the input file. Each test case starts with a line containing an integer N , the number of nodes in a tree, 2<=N<=10,000. The nodes are labeled with integers 1, 2,…, N. Each of the next N -1 lines contains a pair of integers that represent an edge --the first integer is the parent node of the second integer. Note that a tree with N nodes has exactly N - 1 edges. The last line of each test case contains two distinct integers whose nearest common ancestor is to be computed.
Output
Print exactly one line for each test case. The line should contain the integer that is the nearest common ancestor.
Sample Input:
2
16
1 14
8 5
10 16
5 9
4 6
8 4
4 10
1 13
6 15
10 11
6 7
10 2
16 3
8 1
16 12
16 7
5
2 3
3 4
3 1
1 5
3 5
Sample Output:
4
3
题目链接
LCA模板题，适合在线查询最近公共祖先，离线也可以写。
LCA在线AC代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn = 1e4 + 5;
int Rmq[maxn << 1];
struct ST {
int Dp[maxn << 1][20];
void init(int N) {
for (int i = 1; i <= N; ++i) {
Dp[i][0] = i;
}
for (int j = 1; (1 << j) <= 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) {
if (A > B) {
swap(A, B);
}
int K = int(log2(B - A + 1));
return Rmq[Dp[A][K]] <= Rmq[Dp[B - (1 << K) + 1][K]] ? Dp[A][K] : Dp[B - (1 << K) + 1][K];
}
};
struct Link {
int V, Next;
};
Link edges[maxn << 1];
int Head[maxn];
int Tot;
int Vertex[maxn << 1];
int First[maxn];
int Cnt;
ST St;
void Init() {
Tot = 0;
memset(Head, -1, sizeof(Head));
}
void AddEdge(int U, int V) {
edges[Tot] = Link {V, Head[U]};
Head[U] = Tot++;
}
void Dfs(int U, int Pre, int Depth) {
Vertex[++Cnt] = U;
Rmq[Cnt] = Depth;
First[U] = Cnt;
for (int i = Head[U]; i != -1; i = edges[i].Next) {
int V = edges[i].V;
if (V == Pre) {
continue;
}
Dfs(V, U, Depth + 1);
Vertex[++Cnt] = U;
Rmq[Cnt] = Depth;
}
}
void LCA_Init(int Root, int NodeNum) {
Cnt = 0;
Dfs(Root, Root, 0);
St.init(2 * NodeNum - 1);
}
int Query_LCA(int U, int V) {
return Vertex[St.Query(First[U], First[V])];
}
bool Flag[maxn];
int main(int argc, char *argv[]) {
int T;
scanf("%d", &T);
for (int Case = 1, N; Case <= T; ++Case) {
scanf("%d", &N);
Init();
memset(Flag, false, sizeof(Flag));
for (int i = 1, U, V; 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);
int QueryU, QueryV;
scanf("%d%d", &QueryU, &QueryV);
printf("%d\n", Query_LCA(QueryU, QueryV));
}
return 0;
}
LCA离线AC代码:
#include <cstdio>
#include <iostream>
#include <vector>
#include <cstring>
using namespace std;
const int maxn = 1e4 + 5;
struct Edge {
int V, Next;
};
struct Query {
int Q, Next;
int Index;
};
int Pre[maxn << 2];
Edge edges[maxn << 2];
int Head[maxn];
int Tot;
Query querys[maxn << 2];
int Answer[maxn];
int QHead[maxn];
int QTot;
int Vis[maxn];
int Ancestor[maxn];
int F[maxn];
int Find(int X) {
int R = X;
while (Pre[R] != -1) {
R = Pre[R];
}
return R;
}
void Join(int U, int V) {
int RU = Find(U);
int RV = Find(V);
if (RU != RV) {
Pre[RU] = RV;
}
}
void AddEdge(int U, int V) {
edges[Tot] = Edge {V, Head[U]};
Head[U] = Tot++;
}
void AddQuery(int U, int V, int Index) {
querys[QTot] = Query {V, QHead[U], Index};
QHead[U] = QTot++;
querys[QTot] = Query {U, QHead[V], Index};
QHead[V] = QTot++;
}
void Init() {
Tot = 0;
memset(Head, -1, sizeof(Head));
QTot = 0;
memset(QHead, -1, sizeof(QHead));
memset(Vis, false, sizeof(Vis));
memset(Pre, -1, sizeof(Pre));
memset(Ancestor, 0, sizeof(Ancestor));
memset(F, -1, sizeof(F));
}
void LCA(int Node) {
Ancestor[Node] = Node;
Vis[Node] = true;
for (int i = Head[Node]; i != -1; i = edges[i].Next) {
if (Vis[edges[i].V]) {
continue;
}
LCA(edges[i].V);
Join(Node, edges[i].V);
Ancestor[Find(Node)] = Node;
}
for (int i = QHead[Node]; i != -1; i = querys[i].Next) {
if (Vis[querys[i].Q]) {
Answer[querys[i].Index] = Ancestor[Find(querys[i].Q)];
}
}
}
int main(int argc, char *argv[]) {
int T;
scanf("%d", &T);
for (int Case = 1, N; Case <= T; ++Case) {
Init();
scanf("%d", &N);
for (int i = 1, U, V; i < N; ++i) {
scanf("%d%d", &U, &V);
AddEdge(U, V);
F[V] = U;
}
for (int i = 1, U, V; i <= 1; ++i) {
scanf("%d%d", &U, &V);
AddQuery(U, V, i);
}
int Root = 1;
while (F[Root] != -1) {
Root = F[Root];
}
LCA(Root);
printf("%d\n", Answer[1]);
}
return 0;
}
