POJ 1330 Nearest Common Ancestors（倍增）

题目

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

 

题解

 倍增模版题

 

AC代码

#include <iostream>
#include <algorithm>
using namespace std;

const int MAX = 10000 + 10;
const int LOG = 16;
int cnt = 0;
int head[MAX];
bool vis[MAX]; //找root

//-------存边----------
struct Edge{
    int to,next;
}edge[MAX];

void addedge(int u, int v){
    edge[cnt].next = head[u]; //存的是在next边在edge中的下标
    edge[cnt].to = v;  //to指向点
    head[u] = cnt++;
}

//-----初始化anc[x][0]和深度-------
int dep[MAX];
int anc[MAX][LOG];
void dfs(int u, int p, int d){ //p代表父节点
    anc[u][0] = p; dep[u] = d;
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v= edge[i].to;
        //if(v == p) continue; 这里不可能,因为题目里已经给明父子关系，但是其他地方有用
        dfs(v, u, d + 1);
    }
}

//----初始化anc[x][j]----------
void anc_init(int n){  //n为点的个数
    for(int j = 1; j < LOG; ++j)
        for(int i = 1; i <= n; ++i) //将所有的点跳跃2^j步的结果算完后，再接着算2^(j+1)的情况
            anc[i][j] = anc[ anc[i][j - 1] ][j - 1];
}

//----使两点高度相等---
void up(int &u, int h){
    for(int j = 0; h; ++j){
        if(h&1)
            u = anc[u][j];
        h >>= 1;
    }
}


//----LCA--------
int LCA(int u, int v){
    if(dep[u] < dep[v]) swap(u, v); //置u为较深点
    up(u, dep[u] - dep[v]);
    if(u == v) return u;
    for(int i  = LOG - 1; i >= 0; --i){
        if(anc[u][i] != anc[v][i]){
            u = anc[u][i];
            v = anc[v][i];
        }
    }
    return anc[u][0];
}


int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T; cin >> T;
    int N;
    while(T--){
        cin >> N;
        int f, s;
        cnt = 0;
        fill(head, head + MAX, -1); //通过head数组访问对应边在edge中的下标
        fill(vis, vis + MAX, true);
        for(int i = 0; i < N - 1; ++i){
            cin >> f >> s;
            addedge(f, s);
            vis[s] = false;
        }
        int root;
        for(int i = 1; i <= N; ++i)
            if(vis[i]){
                root = i;
                break;
            }
        dfs(root, root, 1); //？？
        anc_init(N);
        int a, b; cin >> a >> b;
        cout << LCA(a, b) << endl;
    }
    return 0;
}