【POJ1330】Nearest Common Ancestors（树链剖分求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

【题意】
就是求两个点的最近公共祖先。

【分析】
　　这题可以用倍增做，这次我用了树剖，感觉还挺好打的，嗯嗯...
　　感觉是因为不可能跳两条重链都越过LCA，因为一个点向下只连一条重边。所以每次调dep较大的边跳就好了。

int LCA(int a, int b)
{
    while (1)
    {
        if(top[a]==top[b]) return dep[a]<=dep[b]?a:b;
        else if(dep[top[a]]>=dep[top[b]]) a=fa[top[a]];
        else b=fa[top[b]];
    }
}

　　就是这样。

　　感觉是因为不可能跳两条边都越过LCA的，因为一个点向下只连着一条重边，所以我们每次选短的那一条跳就好了。

　　http://www.xuebuyuan.com/552070.html

　　这里有详细证明^^^

代码如下：

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 #define Maxn 10010
 #define INF 100000000

 int fa[Maxn],first[Maxn],size[Maxn],dep[Maxn],son[Maxn];
 int w[Maxn],top[Maxn];int wl;
 bool q[Maxn];

 struct node
 {
     int x,y,next;
 }t[*Maxn];int len;

 int mymax(int x,int y) {return x>y?x:y;}
 int mymin(int x,int y) {return x<y?x:y;}

 void ins(int x,int y)
 {
     t[++len].x=x;t[len].y=y;
     t[len].next=first[x];first[x]=len;
 }

 void dfs1(int x,int f)
 {
     fa[x]=f;dep[x]=dep[f]+;size[x]=;
     son[x]=;
     for(int i=first[x];i;i=t[i].next) if(t[i].y!=f)
     {
         dfs1(t[i].y,x);
         size[x]+=size[t[i].y];
         if(size[t[i].y]>size[son[x]]) son[x]=t[i].y;
     }
 }

 void dfs2(int x,int tp)
 {
     w[x]=++wl;
     top[x]=tp;
     if(size[x]!=) dfs2(son[x],tp);
     for(int i=first[x];i;i=t[i].next) if(t[i].y!=fa[x]&&t[i].y!=son[x])
     {
         dfs2(t[i].y,t[i].y);
     }
 }

 int LCA(int a, int b)
 {
     while ()
     {
         if(top[a]==top[b]) return dep[a]<=dep[b]?a:b;
         else if(dep[top[a]]>=dep[top[b]]) a=fa[top[a]];
         else b=fa[top[b]];
     }
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     while(T--)
     {
         int n;
         scanf("%d",&n);
         memset(first,,sizeof(first));
         memset(q,,sizeof(q));
         len=;
         for(int i=;i<n;i++)
         {
             int x,y,c;
             scanf("%d%d",&x,&y);
             q[y]=;
             ins(x,y);//ins(y,x);
         }
         int root;
         for(int i=;i<=n;i++) if(!q[i]) root=i;
         dep[]=;size[]=;
         dfs1(root,);wl=;
         dfs2(root,);
         int x,y;
         scanf("%d%d",&x,&y);
         printf("%d\n",LCA(x,y));
     }
     return ;
 }

[POJ1330]

2016-05-08 17:13:07