lca入门———树上倍增法（博文内含例题）

倍增求LCA：
father【i】【j】表示节点i往上跳2^j次后的节点
可以转移为

father【i】【j】=father【father【i】【j-1】】【j-1】

整体思路：
先比较两个点的深度，如果深度不同，先让深的点往上跳，浅的先不动，等两个点深度一样时，if 相同 直接返回，if 不同 进行下一步；如果不同，两个点一起跳，j从大到小枚举（其实并不大），如果两个点都跳这么多后，得到的点相等，两个点都不动（因为有可能正好是LCA也有可能在LCA上方），知道得到的点不同，就可以跳上来，然后不断跳，两个点都在LCA下面那层，所以再跳1步即可，当father【i】【j】中j=0时即可，就是LCA，返回值结束

（摘自http://blog.csdn.net/dad3zz）

在网络海洋中搜寻很久后找到hzwer的宽搜求deep序，进行了学习。

以下是我的模板代码：

构造访问数组！

void make_bin()
{
    bin[0]=1;
    for(int i=1;i<=15;i++)
        bin[i]=bin[i-1]<<1;
}

简明易懂邻接表！

void Link()
{
    for(int i=1;i<=n-1;i++){//n-1 edges;or you can define a m pointing to the number of edges;
        scanf("%d%d",&u[i],&v[i]);
        u[i+n-1]=v[i];v[i+n-1]=u[i];//双向;
        next[i]=first[u[i]];
        next[i+n-1]=first[v[i]];//双向;
        first[u[i]]=i;
        first[v[i]]=i+n-1;//双向;
        isroot[v[i]]=true;
    }
}

宽搜！

void bfs()
{
    int head=0,tail=1;
    q[0]=root,vis[root]=true;
    while(head^tail){
        int now=q[head];head++;
        for(int i=1;i<=15;i++){
            if(bin[i]<=deep[now])
                fa[now][i]=fa[fa[now][i-1]][i-1];
            else break;
        }
        for(int i=first[now];i;i=next[i])
            if(!vis[v[i]]){
                vis[v[i]]=true;
                fa[v[i]][0]=now;
                deep[v[i]]=deep[now]+1;
                q[tail++]=v[i];
            }
    }
}

求lca()

int lca(int x,int y)
{
    if(deep[x]<deep[y])swap(x,y);
    int t=deep[x]-deep[y];
    for(int i=0;i<=15;i++)
        if(t&bin[i])x=fa[x][i];
    for(int i=15;i>=0;i--)
        if(fa[x][i]^fa[y][i])
            x=fa[x][i],y=fa[y][i];
    if(!(x^y))return y;
    return fa[x][0];
}

之后，在poj上找了一个众所周知的例题，poj1330，当做lca入门题

Nearest Common Ancestors

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24757		Accepted: 12864

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

Source

Taejon 2002

Source Code

Problem: 1330		User: ksq2013
Memory: 1224K		Time: 32MS
Language: C++		Result: Accepted

ac代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
bool vis[10100],isroot[10100];
int root,q[10100],deep[10100],fa[10100][20];
int n,bin[20],first[10100],next[10100],u[10100],v[10100];
void make_bin()
{
    bin[0]=1;
    for(int i=1;i<=15;i++)
        bin[i]=bin[i-1]<<1;
}
void Link()
{
    for(int i=1;i<=n-1;i++){
        scanf("%d%d",&u[i],&v[i]);
        //u[i+n-1]=v[i];v[i+n-1]=u[i];
        next[i]=first[u[i]];
        //next[i+n-1]=first[v[i]];
        first[u[i]]=i;
        //first[v[i]]=i+n-1;
        isroot[v[i]]=true;
    }
    for(int i=1;i<=n;i++)
        if(!isroot[i]){
            root=i;
            break;
        }
}
void bfs()
{
    int head=0,tail=1;
    q[0]=root,vis[root]=true;
    while(head^tail){
        int now=q[head];head++;
        for(int i=1;i<=15;i++){
            if(bin[i]<=deep[now])
                fa[now][i]=fa[fa[now][i-1]][i-1];
            else break;
        }
        for(int i=first[now];i;i=next[i])
            if(!vis[v[i]]){
                vis[v[i]]=true;
                fa[v[i]][0]=now;
                deep[v[i]]=deep[now]+1;
                q[tail++]=v[i];
            }
    }
}
int lca(int x,int y)
{
    if(deep[x]<deep[y])swap(x,y);
    int t=deep[x]-deep[y];
    for(int i=0;i<=15;i++)
        if(t&bin[i])x=fa[x][i];
    for(int i=15;i>=0;i--)
        if(fa[x][i]^fa[y][i])
            x=fa[x][i],y=fa[y][i];
    if(!(x^y))return y;
    return fa[x][0];
}
int main()
{
    make_bin();
    int T;
    scanf("%d",&T);
    for(;T;T--){
        memset(vis,false,sizeof(vis));
        memset(isroot,false,sizeof(isroot));
        memset(q,0,sizeof(q));
        memset(fa,0,sizeof(fa));
        memset(deep,0,sizeof(deep));
        memset(first,0,sizeof(first));
        memset(next,0,sizeof(next));
        scanf("%d",&n);
        Link();
        bfs();
        int x,y;
        scanf("%d%d",&x,&y);
        printf("%d
",lca(x,y));
    }
    return 0;
}