Closest Common Ancestors
Time Limit: 2000MS   Memory Limit: 10000K
Total Submissions: 13370   Accepted: 4338

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

模板题

 /* ***********************************************
Author :kuangbin
Created Time :2013-9-5 8:54:16
File Name :F:\2013ACM练习\专题学习\LCA\POJ1470.cpp
************************************************ */ #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = ;
int rmq[*MAXN];//rmq数组,就是欧拉序列对应的深度序列
struct ST
{
int mm[*MAXN];
int dp[*MAXN][];//最小值对应的下标
void init(int n)
{
mm[] = -;
for(int i = ;i <= n;i++)
{
mm[i] = ((i&(i-)) == )?mm[i-]+:mm[i-];
dp[i][] = i;
}
for(int j = ; j <= mm[n];j++)
for(int i = ; i + (<<j) - <= n; i++)
dp[i][j] = rmq[dp[i][j-]] < rmq[dp[i+(<<(j-))][j-]]?dp[i][j-]:dp[i+(<<(j-))][j-];
}
int query(int a,int b)//查询[a,b]之间最小值的下标
{
if(a > b)swap(a,b);
int k = mm[b-a+];
return rmq[dp[a][k]] <= rmq[dp[b-(<<k)+][k]]?dp[a][k]:dp[b-(<<k)+][k];
}
};
//边的结构体定义
struct Edge
{
int to,next;
};
Edge edge[MAXN*];
int tot,head[MAXN]; int F[MAXN*];//欧拉序列,就是dfs遍历的顺序,长度为2*n-1,下标从1开始
int P[MAXN];//P[i]表示点i在F中第一次出现的位置
int cnt; ST st;
void init()
{
tot = ;
memset(head,-,sizeof(head));
}
void addedge(int u,int v)//加边,无向边需要加两次
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void dfs(int u,int pre,int dep)
{
F[++cnt] = u;
rmq[cnt] = dep;
P[u] = cnt;
for(int i = head[u];i != -;i = edge[i].next)
{
int v = edge[i].to;
if(v == pre)continue;
dfs(v,u,dep+);
F[++cnt] = u;
rmq[cnt] = dep;
}
}
void LCA_init(int root,int node_num)//查询LCA前的初始化
{
cnt = ;
dfs(root,root,);
st.init(*node_num-);
}
int query_lca(int u,int v)//查询u,v的lca编号
{
return F[st.query(P[u],P[v])];
}
bool flag[MAXN];
int Count_num[MAXN];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
int u,v,k;
int Q;
while(scanf("%d",&n) == )
{
init();
memset(flag,false,sizeof(flag));
for(int i = ;i <= n;i++)
{
scanf("%d:(%d)",&u,&k);
while(k--)
{
scanf("%d",&v);
flag[v] = true;
addedge(u,v);
addedge(v,u);
}
}
int root;
for(int i = ;i <= n;i++)
if(!flag[i])
{
root = i;
break;
}
LCA_init(root,n);
memset(Count_num,,sizeof(Count_num));
scanf("%d",&Q);
while(Q--)
{
char ch;
cin>>ch;
scanf("%d %d)",&u,&v);
Count_num[query_lca(u,v)]++;
}
for(int i = ;i <= n;i++)
if(Count_num[i] > )
printf("%d:%d\n",i,Count_num[i]);
}
return ;
}

POJ 1470 Closest Common Ancestors (LCA, dfs+ST在线算法)的更多相关文章

  1. POJ - 1330 Nearest Common Ancestors(dfs+ST在线算法|LCA倍增法)

    1.输入树中的节点数N,输入树中的N-1条边.最后输入2个点,输出它们的最近公共祖先. 2.裸的最近公共祖先. 3. dfs+ST在线算法: /* LCA(POJ 1330) 在线算法 DFS+ST ...

  2. POJ 1330 Nearest Common Ancestors (dfs+ST在线算法)

    详细讲解见:https://blog.csdn.net/liangzhaoyang1/article/details/52549822 zz:https://www.cnblogs.com/kuang ...

  3. poj 1470 Closest Common Ancestors LCA

    题目链接:http://poj.org/problem?id=1470 Write a program that takes as input a rooted tree and a list of ...

  4. POJ 1470 Closest Common Ancestors(LCA&RMQ)

    题意比较费劲:输入看起来很麻烦.处理括号冒号的时候是用%1s就可以.还有就是注意它有根节点...Q次查询 在线st算法 /*************************************** ...

  5. POJ 1470 Closest Common Ancestors(LCA 最近公共祖先)

    其实这是一个裸求LCA的题目,我使用的是离线的Tarjan算法,但是这个题的AC对于我来说却很坎坷……首先是RE,我立马想到数组开小了,然后扩大了数组,MLE了……接着把数组调整适当大小,又交了一发, ...

  6. POJ 1470 Closest Common Ancestors LCA题解

    本题也是找LCA的题目,只是要求多次查询.一般的暴力查询就必定超时了,故此必须使用更高级的方法,这里使用Tarjan算法. 本题处理Tarjan算法,似乎输入处理也挺麻烦的. 注意: 由于查询的数据会 ...

  7. POJ 1470 Closest Common Ancestors(最近公共祖先 LCA)

    POJ 1470 Closest Common Ancestors(最近公共祖先 LCA) Description Write a program that takes as input a root ...

  8. POJ 1470 Closest Common Ancestors 【LCA】

    任意门:http://poj.org/problem?id=1470 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000 ...

  9. POJ 1470 Closest Common Ancestors (LCA,离线Tarjan算法)

    Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 13372   Accept ...

随机推荐

  1. Linux下获取和设置IP

    在Linux下获取关于IP和网关的操作:重点是对struct ifreq 的操作. 那么进入目录/usr/include/net/if.h下看查找struct ifreq结构体. /* Interfa ...

  2. Python元组与字典详解

    Python 元组 Python的元组与列表类似,不同之处在于元组的元素不能修改. 元组使用小括号,列表使用方括号. 元组创建很简单,只需要在括号中添加元素,并使用逗号隔开即可. 如下实例: tup ...

  3. (四)Jsoup 获取 DOM 元素属性值

    第一节: Jsoup 获取 DOM 元素属性值 Jsoup获取DOM元素属性值 比如我们要获取博客的href属性值: 我们这时候就要用到Jsoup来获取属性的值 : 我们给下示例代码: package ...

  4. Ubuntu CEPH快速安装

    一.CEPH简介 不管你是想为云平台提供Ceph 对象存储和/或 Ceph 块设备,还是想部署一个 Ceph 文件系统或者把 Ceph 作为他用,所有 Ceph 存储集群的部署都始于部署一个个 Cep ...

  5. Python装饰器讲解

    Python装饰器讲解 定义:本质是函数,就是为其他函数添加附加功能.原则:1.不能修改被装饰的函数的源代码 2.不能修改被装饰的函数的调用方式 import time def timmer(func ...

  6. 多线程-TaskExecutor-使用Demo

    BasicExecute.java package com.jef.executeTest; public abstract class BasicExecute extends Thread { @ ...

  7. echarts3.0 本期累计堆叠

    @{ ViewBag.Title = "barlj"; } <h2>barlj</h2> <div id="main" style ...

  8. 使用scss + react + webpack + es6实现幻灯片

    写在前面: 刚学习完慕课网里的一个幻灯片案例,自己加了刚学的react,两者结合.首先让大家看看效果 点击此处 你可以先用纯js实现上面的效果:我的github上的 JS代码 或者 观看慕课提供的课程 ...

  9. 【51nod】1742 开心的小Q

    题解 我们由于莫比乌斯函数如果有平方数因子就是0,那么我们可以列出这样的式子 \(\sum_{i = 1}^{n} \sum_{d|i} (1 - |\mu(d)|)\) 然后枚举倍数 \(\sum_ ...

  10. 005 Hadoop的三种模式区别

    1.本地模式 -默认模式. -不对配置文件进行修改. -使用本地文件系统,而不是分布式文件系统. -Hadoop不会启动NameNode.DataNode.ResourceManager.NodeMa ...