poj 1330 Nearest Common Ancestors(倍增法)

Nearest Common Ancestors

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 14078 Accepted: 7510

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

2161 148 510 165 94 68 44 101 136 1510 116 710 216 38 116 1216 752 33 43 11 53 5

Sample Output

43

Source

Taejon 2002

在线算法 倍增法

每次询问O(logN)

d[i] 表示 i节点的深度, p[i,j] 表示 i 的 2^j 倍祖先

那么就有一个递推式子 p[i,j]=p[p[i,j-1],j-1]

这样子一个O(NlogN)的预处理求出每个节点的 2^k 的祖先

然后对于每一个询问的点对a, b的最近公共祖先就是：

先判断是否 d[a] > d[b] ,如果是的话就交换一下(保证 a 的深度小于 b 方便下面的操作)然后把b 调到与a 同深度, 同深度以后再把a, b 同时往上调(dec(j)) 调到有一个最小的j 满足p[a,j]!=p[b,j] (a b 是在不断更新的), 最后再把 a, b 往上调 (a=p[a,0], b=p[b,0]) 一个一个向上调直到a = b, 这时 a or b 就是他们的最近公共祖先

http://poj.org/problem?id=1330 

【分析】

倍增法求LCA的大体思路：

     首先确定深度d[i]，表示i到根节点的距离或者说是深度。

     p[i,j]表示i向上的第2^j个祖先，则p[i,j+1]:=p[p[i,j],j];

     dfs求出d[i]和p[i,j]。

     如果求的是x和y的lca。那么如果x=y，lca=x or y;

     如果x和y的深度不同，则调整到同一深度，以便求lca。

     x和y的深度相同了，就开始向上走，如果两个点同时走了相同的步数，走到了一个点，那么这个点就是lca，自己想象一下，这是显然的事。两个点最早到达的相同点不是lca是什么= =

     结束。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;#define N 10010#define M 20int d[N],f[N][M];vector<int>ch[N];void dfs(int x){//求出所有结点深度d[x]=d[f[x][0]]+1;for(int i=1;i<M;i++)f[x][i]=f[f[x][i-1]][i-1];//倍增祖先for(int i=ch[x].size()-1;i>=0;i--)dfs(ch[x][i]);//遍历儿子}int lca(int x,int y){//求x,y最小公共祖先    if(d[x]<d[y])swap(x,y);//保证x的深度不小于yint k=d[x]-d[y];for(int i=0;i<M;i++)if((1<<i)&k)x=f[x][i];if(x==y)return x;//x==y时为最小公共祖先for(int i=M-1;i>=0;i--)if(f[x][i]!=f[y][i]){x=f[x][i];y=f[y][i];}return f[x][0];}int main(){int T,n,i,x,y;scanf("%d",&T);while(T--){scanf("%d",&n);memset(d,0,sizeof(d));memset(f,0,sizeof(f));memset(ch,0,sizeof(ch));for(i=1;i<n;i++){scanf("%d%d",&x,&y);ch[x].push_back(y);f[y][0]=x;}for(i=1;i<=n;i++)if(f[i][0]==0){//找到根遍历dfs(i);break;}scanf("%d%d",&x,&y);printf("%d\n",lca(x,y));}return 0;}