POJ1470/ZOJ1141 Closest Common Ancestors（LCA离线算法）

Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and v and has the greatest depth in the tree. A node can be its own ancestor (for example in Figure 1 the ancestors of node 2 are 2 and 5)

Input

The data set, which is read from a the std input, starts with the tree description, in the form: 

nr_of_vertices 
vertex:(nr_of_successors) successor1 successor2 ... successorn 
... 
where vertices are represented as integers from 1 to n ( n <= 900 ). The tree description is followed by a list of pairs of vertices, in the form: 
nr_of_pairs 
(u v) (x y) ... 

The input file contents several data sets (at least one). 
Note that white-spaces (tabs, spaces and line breaks) can be used freely in the input.

Output

For each common ancestor the program prints the ancestor and the number of pair for which it is an ancestor. The results are printed on the standard output on separate lines, in to the ascending order of the vertices, in the format: ancestor:times 
For example, for the following tree: 

Sample Input

5
5:(3) 1 4 2
1:(0)
4:(0)
2:(1) 3
3:(0)
6
(1 5) (1 4) (4 2)
      (2 3)
(1 3) (4 3)

Sample Output

2:1
5:5

Hint

Huge input, scanf is recommended.

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<stack>
#include<vector>
#define nn 1001
#define mm 500001
#define inff 0x3fffffff
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
int fa[nn],n,m,num,num1;
int vis[nn];
int ans[nn];
int sum[nn];
int p[nn],p1[nn];
struct node
{
    int en,next;
}q[mm],e[mm];
int add(int st,int en)
{
    e[num].en=en;e[num].next=p[st];p[st]=num++;
}
int add1(int st,int en)
{
    q[num1].en=en;q[num1].next=p1[st];
    p1[st]=num1++;
}
void init()
{
    memset(vis,0,sizeof(vis));
    memset(ans,0,sizeof(ans));
    memset(sum,0,sizeof(sum));
    memset(p,-1,sizeof(p));
    memset(p1,-1,sizeof(p1));
    num=num1=0;
    for(int i=1;i<=n;i++)
        fa[i]=i;
}
int find(int u)
{
    while(u!=fa[u])
        u=fa[u];
    return u;
}
void dfs(int u)
{
    for(int i=p[u];i+1;i=e[i].next)
    {
        int v=e[i].en;
        dfs(v);
        fa[v]=u;
    }
    for(int i=p1[u];i+1;i=q[i].next)
    {
        int v=q[i].en;
        if(vis[v])
        {
            int aim=find(v);
            ans[aim]++;
        }
    }
    vis[u]=1;
}
int main()
{
    int u,v,t;
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("\t%d\t:\t(\t%d\t)",&u,&t);
            for(int i=1;i<=t;i++)
            {
                scanf("\t%d\t",&v);
                add(u,v);
                sum[v]=1;
            }
        }
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            getchar();
            scanf("\t(%d\t %d\t)",&u,&v);
             add1(u,v);
             add1(v,u);
        }
     int ff;
     for(ff=1;ff<=n;ff++)
        if(!sum[ff]) break;
     dfs(ff);
        for(int i=1;i<=n;i++)
        {
            if(ans[i])
            {
                printf("%d:%d\n",i,ans[i]);
            }
        }
    }
}