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. WCF - Autofac IOC

    /// <summary> /// IOC实例提供者,基于AutoFac /// /// </summary> public class IocInstanceProvider ...

  2. opencv(4)实现数据增加小工具

    数据增加(data augmentation),作为一种深度学习中的常用手段,数据增加对模型的泛化性和准确性都有帮助.数据增加的具体使用方式一般有两种,一种是实时增加,比如在Caffe中加入数据扰动层 ...

  3. Vue select 下拉菜单

    1.html <div id="app-8"> <select v-model="selected"> <option v-for ...

  4. bzoj [SDOI2009]学校食堂Dining

    感觉这个状压dp比较难想.. dp[ i ][ s ][ k ] 表示前i - 1个都排好了, 从i开始的7个的取没取的状态为s, 且最后一个相对i的位置为k的最少花费. 状态转移方程 if(s &a ...

  5. hdoj2037 今年暑假不AC(贪心)

    题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=2037 思路 想要看的节目尽可能的多,则首先要将节目按照结束时间从早到晚排序,因为一个节目越早结束,留给 ...

  6. CentOS源码安装搭建LNMP全过程(包括nginx,mysql,php,svn)

    服务器环境为:CentOS6.5 64位 目标:搭建LNMP(Linux + Nginx + MySQL + PHP +SVN),其中svn是用来代替ftp,方便开发中调试同步代码 相关目录:所有软件 ...

  7. 基于Laravel开发博客应用系列 —— 十分钟搭建博客系统

    1.创建文章数据表及其模型(0:00~2:30) 我们已经在上一节中为博客项目完成了大部分准备工作,现在首先要做的就是为这个项目创建一个新的文章表 posts及该表对应的模型类 Post,使用如下Ar ...

  8. Java_集合与泛型

    Collection 集合,集合是java中提供的一种容器,可以用来存储多个数据.在前面的学习中,我们知道数据多了,可以使用数组存放或者使用ArrayList集合进行存放数据.那么,集合和数组既然都是 ...

  9. leetcode 入门第一题 4ms? 8ms? Two Sum

    今天开启leetcode 入门第一题 题意很简单,就是一个数组中求取两数之和等于目标数的一对儿下标 1.暴力 n^2 两个for循环遍历 用时0.1s 开外 代码就不用写了 2.二分 nlogn 我们 ...

  10. [代码审计]DM企业建站系统v201710 sql注入漏洞分析 | 新版v201712依旧存在sql注入

    0x00 前言 本来呢,这套CMS都不想审的了.下载下来打开一看,各种debug注释,排版烂的不行. 贴几个页面看看 感觉像是新手练手的,没有审下去的欲望了. 但想了想,我tm就是新手啊,然后就继续看 ...