1470 Closest Common Ancestors //LCA

Closest Common Ancestors

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 7027		Accepted: 2199

Description

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.

Source

Southeastern Europe 2000

 

#include<stdio.h>
#include<vector>
using namespace std;
const int MAX=901;
int f[MAX];
int r[MAX];
int indegree[MAX];//保存每个节点的入度
int visit[MAX];
vector<int> tree[MAX],Qes[MAX];
int ancestor[MAX];
int sum[MAX];
void init(int n)
{
    for(int i=1;i<=n;i++)
    {
        r[i]=1;//类似于rank数组
        f[i]=i;//用于并查集记录父亲节点
        indegree[i]=0;//节点的层号
        visit[i]=0;//是否被访问
        ancestor[i]=0;//祖先
        tree[i].clear();//清空
        Qes[i].clear();
        sum[i]=0;
    }

}
int find(int n)
{  //查找函数，并压缩路径
    if(f[n]==n) return n;
    else f[n]=find(f[n]);
    return f[n];
}
int Union(int x,int y)
{ //合并函数，如果属于同一分支则返回0，成功合并返回1
    int a=find(x);
    int b=find(y);
    if(a==b) return 0;
    //相等的话,x向y合并
    else if(r[a]<=r[b])
    {
        f[a]=b;
        r[b]+=r[a];
    }
    else
    {
        f[b]=a;
        r[a]+=r[b];
    }
    return 1;
}
void LCA(int u)
{
    ancestor[u]=u;
    int size=tree[u].size();
    for(int i=0;i<size;i++)
    {
        LCA(tree[u][i]);
        Union(u,tree[u][i]);
        ancestor[find(u)]=u;
    }
    visit[u]=1;
    size=Qes[u].size();
    for(int i=0;i<size;i++)
    {   //如果已经访问了问题节点,就可以返回结果了.
        if(visit[Qes[u][i]]==1)
        {
            //printf("%d/n",ancestor[find(Qes[u][i])]);
            sum[ancestor[find(Qes[u][i])]]++;
            //return;
        }
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        init(n);
        int s,t,num;
        char a,b,c;
        for(int i=1;i<=n;i++)
        {
            scanf(" %d%c%c%d%c ",&s,&a,&b,&num,&c);
            for(int j=1;j<=num;j++)
            {
                scanf("%d",&t);
                tree[s].push_back(t);
                indegree[t]++;
            }
        }
        scanf("%d ",&num);
        for(int i=1;i<=num;i++)
        {
           scanf(" %c %d %d %c",&a,&s,&t,&b);//这里可以输入多组询问
           Qes[s].push_back(t);//相当于询问两次
           Qes[t].push_back(s);
        }
        for(int i=1;i<=n;i++)
        {   //寻找根节点
            if(indegree[i]==0)
            {
                LCA(i);
                break;
            }
        }
        for(int i=1;i<=n;i++)
        if(sum[i]!=0)
        printf("%d:%d/n",i,sum[i]);
    }
    return 0;
}