poj 1470(LCA模板题)

POJ 1470 最近公共祖先(LCA)问题解析

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这篇博客主要讲解了POJ 1470题目，涉及最近公共祖先（Lowest Common Ancestor, LCA）的解决策略。通过输入输出样例分析，阐述了如何利用树形结构处理此类问题，并提供了C语言的解决方案。对于初学者，此模板题有助于理解LCA问题的解决思路。" 28976159,3357679,cocos2dx3.1在Windows上的安装与Android编译教程,"['cocos2dx', '移动开发', 'Android开发', '游戏开发']

Closest Common Ancestors

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 9221		Accepted: 2955

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

题目： http://poj.org/problem?id=1470

分析：还是很明显的LCA问题，数据比较大，测试模板用，随便加个读入优化就79ms。。。没加居然780++ms，囧。。。那些没加优化的100++ms怎么搞来的，不会阿～～～

代码：

#include<cstdio>
using namespace std;
const int mm=1111111;
const int mn=1111;
int t[mm],p[mm];
int h[mn],q[mn],f[mn],sum[mn];
bool vis[mn];
int i,j,k,c,e,n;
int find(int u)
{
    if(f[u]==u)return u;
    else return f[u]=find(f[u]);
}
bool tarjan(int u)
{
    int i,v;
    for(i=q[u];i;i=p[i])
        if(vis[v=t[i]])++sum[find(v)];
    vis[f[u]=u]=1;
    for(i=h[u];i;i=p[i])
        if(!vis[v=t[i]])tarjan(v),f[v]=u;
    return 0;
}
inline bool get(int &a)
{
    char c;
    while(((c=getchar())<'0'||c>'9')&&c!=EOF);
    if(c==EOF)return 0;
    for(a=0;c>='0'&&c<='9';c=getchar())a=a*10+c-'0';
    return 1;
}
int main()
{
    while(get(n))
    {
        for(i=1;i<=n;++i)sum[i]=h[i]=q[i]=f[i]=vis[i]=0;
        for(c=e=1;c<=n;++c)
        {
            get(i),get(k);
            while(k--)get(j),++f[t[e]=j],p[e]=h[i],h[i]=e++;
        }
        get(k);
        while(k--)
        {
            get(i),get(j);
            t[e]=j,p[e]=q[i],q[i]=e++;
            t[e]=i,p[e]=q[j],q[j]=e++;
        }
        for(i=1;i<=n;++i)
            if(!f[i])tarjan(i);
        for(i=1;i<=n;++i)
            if(sum[i])printf("%d:%d\n",i,sum[i]);
    }
    return 0;
}