POJ：1330-Nearest Common Ancestors（LCA在线、离线、优化算法）

传送门：http://poj.org/problem?id=1330

Nearest Common Ancestors

Time Limit: 1000MS Memory Limit: 10000K

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

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解题心得：

1. 这是一个LCA的裸题，可以用用离线算法，也就是tarjan来做这个题，运用并查集，一棵树上的点分为三种，一种是已经找过的点，一种是正在找的点，还有一种是没有找过的点，如果其中一个点是没有找过的就不管，继续找下去，如果一个点是找过的，就直接用并查集回到他们共同的根节点（两个点必然在同一棵子树下），如果两个点都是正在找的点，那么其中一个点就是其最近祖先。
2. 还有一种就是使用倍增法来做这个题，使用倍增法需要知道当前点的深度和父节点。两个点的最近公共祖先就是他们一起向上走第一次遇到的地方，运用这个性质就可以先将两个点的深度调到相同，然后一起向上走，第一次遇到的地方就是其最近公共祖先。
3. 优化的思想也很简单，两个点一步一步的向上走是不是太慢了，可不可以多走几步，那怎么走呢？就可以想到使用RMQ的思想，按照二进制来走，在统一深度的时候可以将深度大的的那个点，走深度小的那个点的二进制中没有1的位置。然后一起向上面走二进制的步数，找打到第一个不是公共祖先的点然后返回他的父节点。思想比较简单，还是看实现过程吧。

---

tarjan写法：

#include<cstring>
#include<stdio.h>
#include<vector>
using namespace std;
const int maxn = 1e4+100;
int father[maxn],q1,q2,ans,n;
bool vis[maxn];
vector <int> ve[maxn];

void init()
{
    memset(vis,0,sizeof(vis));
    memset(father,0,sizeof(father));
    for(int i=0;i<maxn;i++)
        ve[i].clear();
    for(int i=1;i<n;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        ve[a].push_back(b);
        vis[b] = true;
    }
    scanf("%d%d",&q1,&q2);
}

int find(int x)
{
    if(x == father[x])
        return x;
    return father[x] = find(father[x]);
}

void tarjan(int x)
{
    father[x] = x;
    for(int i=0;i<ve[x].size();i++)
    {
        int v = ve[x][i];
        tarjan(v);
        father[v] = x;
    }
    if(x == q1 || x == q2)
    {
        if(x != q1)
            swap(q1,q2);
        if(father[q2])//如果其中一个点没被找到那么就继续找下去
            ans = find(father[q2]);
    }
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<=n;i++)
        {
            if(!vis[i])
            {
                tarjan(i);
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

---

倍增写法(无优化）

#include<stdio.h>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 1e4+100;
vector <int> ve[maxn];
int n,father[maxn],dep[maxn];
bool vis[maxn];

void init()
{
    for(int i=0;i<=n;i++)
        ve[i].clear();
    memset(father,0,sizeof(father));
    memset(dep,0,sizeof(dep));
    memset(vis,0,sizeof(vis));
    for(int i=0;i<n-1;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        vis[b] = true;
        ve[a].push_back(b);
    }
}

int dfs(int u,int f,int d)
{
    father[u] = f;//记录父节点
    dep[u] = d;//记录深度
    for(int i=0;i<ve[u].size();i++)
    {
        int v = ve[u][i];
        if(v == f)//主要是处理单向边
            continue;
        dfs(v,u,d+1);
    }
}

void LCA(int q,int p)
{
    if(dep[p] > dep[q])
        swap(q,p);
    while(dep[q] > dep[p])//调节到同一深度
        q = father[q];
    while(p != q)//一起向上走
    {
        p = father[p];
        q = father[q];
    }
    printf("%d\n",q);
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<=n;i++)
            if(!vis[i])
            {
                dfs(i,-1,0);
                break;
            }
        int q,p;
        scanf("%d%d",&q,&p);
        LCA(q,p);
    }
}

---

LCA进过优化后的代码：

#include<stdio.h>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 1e4+100;
const int LOG = 33;
int p[maxn][33],dep[maxn],n;
bool vis[maxn];
vector <int> ve[maxn];

void init()
{
    for(int i=0;i<=n;i++)
        ve[i].clear();
    memset(dep,0,sizeof(dep));
    memset(p,0,sizeof(p));
    memset(vis,0,sizeof(vis));
    for(int i=1;i<n;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        ve[a].push_back(b);
        vis[b] = true;
    }
}

void dfs(int u,int f,int d)
{
    p[u][0] = f;//u点向前移动2的0次方位为它的父节点
    dep[u] = d;
    for(int i=0;i<ve[u].size();i++)
    {
        int v = ve[u][i];
        if(v == f)
            continue;
        dfs(v,u,d+1);
    }
}

int LCA(int x,int y)
{
    for(int i=0;i+1<LOG;i++)
        for(int j=1;j<=n;j++)
            if(p[j][i] < 0) p[j][i+1] = -1;//树中的节点向上移动超出了根节点都为-1
            else p[j][i+1] = p[p[j][i]][i];//否则RMQ思想
    if(dep[y] > dep[x])
        swap(y,x);
    for(int i=0;i<LOG;i++)
        if(dep[x] - dep[y] >> i & 1)//向上移动到同一深度的时候，将更深的节点二进制表示中多出部分的1移走就行了
            x = p[x][i];
    if(x == y)//同一深度的时候已经合一了
        return x;
    for(int i=LOG-1;i>=0;i--)//找到向上移动中最大祖先的下面第一个节点
    {
        if(p[x][i] != p[y][i])
        {
            x = p[x][i];
            y = p[y][i];
        }
    }
    return p[x][0];
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        init();
        for(int i=1;i<=n;i++)
            if(!vis[i])
            {
                dfs(i,-1,0);
                break;
            }
        int p,q;
        scanf("%d%d",&p,&q);
        printf("%d\n",LCA(p,q));
    }
}