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. Oracle学习笔记:wm_concat函数合并字段

    在Oracle中使用wm_concat(column)可以实现字段的分组合并,逗号分隔. 例如,现有表temp_cwh_test: -- 创建临时表 create table temp_cwh_tes ...

  2. grail开发环境的搭建

    本文参考:Grails入门指南(第二版) 1. 下载jdk和Grail http://www.oracle.com/technetwork/java/javase/downloads/ http:// ...

  3. mybatis中多条件判断---choose when的用法

    <select id="getFunctionByPage" resultMap="FunctionRlt"> SELECT K.FUNCTION_ ...

  4. CentOS6.5配置rsyslog

    如何在RHEL 6.5安装和配置rsyslog现在7.6版本/ CentOS的6.5 .The情况是,安装和RHEL / CentOS的6.5安装rsyslog现在集中式日志服务器上.所有的客户端服务 ...

  5. git/github 生成密钥

    当从本地提交文件到github的时候,提交不成功,报错,可能问题就是你还没有生成ssh秘钥 github要使用ssh密钥的原因: git使用https协议,每次pull, push都要输入密码,相当的 ...

  6. HBase(八)HBase的协处理器

    一.协处理器简介 1. 起源 Hbase 作为列族数据库最经常被人诟病的特性包括:无法轻易建立“二级索引”,难以执 行求和.计数.排序等操作.比如,在旧版本的(<0.92)Hbase 中,统计数 ...

  7. 【PAT】1015 德才论 (25)(25 分)

    1015 德才论 (25)(25 分) 宋代史学家司马光在<资治通鉴>中有一段著名的“德才论”:“是故才德全尽谓之圣人,才德兼亡谓之愚人,德胜才谓之君子,才胜德谓之小人.凡取人之术,苟不得 ...

  8. 【LOJ】#2722. 「NOI2018」情报中心

    https://loj.ac/problem/2722 题解 考场上想了60分,但是由于自己不知道在怎么zz,我连那个ai<bi都没看到,误以为出题人没给lca不相同的部分分,然后觉得lca不同 ...

  9. Python全栈开发之16、jquery

    一.查找元素 1.选择器 1.1 基本选择器      $("*")  $("#id")  $(".class")  $("ele ...

  10. PHP学习日记 Windows配置PHP+Nginx+自动化脚本

    Windows配置PHP+Nginx+自动化脚本 安装与配置 PHP 下载PHP:传送门 选择合适的版本下载 尽量选Thread Safe 配置PHP: 解压后在文件夹中找到php.ini-devel ...