Closest Common Ancestors
Time Limit: 2000MS   Memory Limit: 10000K
Total Submissions: 20804   Accepted: 6608

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

 
代码:

#include<cstdio>
#include<vector>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 10100
using namespace std;
char ch;
vector<int>vec[N],que[N];
int t,s,n,m,x,y,num,qx[N],qy[N],fa[N],dad[N],ans[N],root,ans1[N];
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
int find(int x)
{
    if(fa[x]==x) return x;
    fa[x]=find(fa[x]);
    return fa[x];
}
int tarjan(int x)
{
    fa[x]=x;
    ;i<vec[x].size();i++)
     if(vec[x][i]!=dad[x])
      dad[vec[x][i]]=x,tarjan(vec[x][i]);
    ;i<que[x].size();i++)
     if(dad[y=qx[que[x][i]]^qy[que[x][i]]^x])
      ans1[que[x][i]]=find(y);
    fa[x]=dad[x];
}
void begin()
{
    ;i<=n;i++)
     vec[i].clear(),que[i].clear();
    memset(fa,,sizeof(fa));
    memset(ans,,sizeof(ans));
    memset(dad,,sizeof(dad));
    memset(ans1,,sizeof(ans1));
}
int main()
{
    while(scanf("%d",&t)!=EOF)
    {
        s=t;begin();
        while(t--)
        {
            x=read();
            n=read();
            ;i<=n;i++)
            {
                y=read();fa[y]=x;
                vec[x].push_back(y);
                vec[y].push_back(x);
             }
        }
        ;i<=s;i++)
         if(!fa[i]) root=i;
        memset(fa,,sizeof(fa));
        memset(ans,,sizeof(ans));
        m=read();
        ;i<=m;i++)
        {
            qx[i]=read(),qy[i]=read();
            que[qx[i]].push_back(i);
            que[qy[i]].push_back(i);
        }
        tarjan(root);
        ;i<=m;i++)
          ans[ans1[i]]++;
        ;i<=s;i++)
         if(ans[i]) printf("%d:%d\n",i,ans[i]);
    }
    ;
}

tarjan暴空间、、、

O(≧口≦)O气死了,蒟蒻表示以后再也不用tarjan了!!!!!!!!!!!!

#include<vector>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define N 910
using namespace std;
vector<int>vec[N];
int n,m,s,x,y,dad[N],fa[N],top[N],deep[N],size[N],ans[N];
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
int lca(int x,int y)
{
    for(;top[x]!=top[y];)
    {
        if(deep[top[x]]<deep[top[y]])
         swap(x,y);
        x=fa[x];
    }
    if(deep[x]>deep[y])
     swap(x,y);
    return x;
}
int dfs(int x)
{
    size[x]=;
    deep[x]=deep[fa[x]]+;
    ;i<vec[x].size();i++)
     if(vec[x][i]!=fa[x])
    {
        fa[vec[x][i]]=x;
        dfs(vec[x][i]);
        size[x]+=size[vec[x][i]];
    }
}
int dfs1(int x)
{
    ;
    if(!top[x]) top[x]=x;
    ;i<vec[x].size();i++)
     if(vec[x][i]!=fa[x]&&size[t]<size[vec[x][i]])
      t=vec[x][i];
    if(t) top[t]=top[x],dfs1(t);
    ;i<vec[x].size();i++)
     if(vec[x][i]!=fa[x]&&vec[x][i]!=t)
      dfs1(vec[x][i]);
}
int begin()
{
    ;i<=n;i++)
      vec[i].clear();
    memset(fa,,sizeof(fa));
    memset(top,,sizeof(top));
    memset(ans,,sizeof(ans));
    memset(dad,,sizeof(dad));
    memset(deep,,sizeof(deep));
    memset(size,,sizeof(size));
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        s=n;begin();
        while(n--)
        {
            x=read();m=read();
            ;i<=m;i++)
            {
                y=read();dad[y]=x;
                vec[x].push_back(y);
                vec[y].push_back(x);
            }
        }
        ;i<=s;i++)
         if(!dad[i])
          {dfs(i);dfs1(i);break;}
        m=read();
        ;i<=m;i++)
        {
            x=read(),y=read();
            ans[lca(x,y)]++;
        }
        ;i<=s;i++)
         if(ans[i]) printf("%d:%d\n",i,ans[i]);
    }
    ;
}

poj——1470 Closest Common Ancestors的更多相关文章

  1. 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 ...

  2. POJ 1470 Closest Common Ancestors 【LCA】

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

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

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

  4. POJ 1470 Closest Common Ancestors

    传送门 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 17306   Ac ...

  5. POJ 1470 Closest Common Ancestors (LCA, dfs+ST在线算法)

    Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 13370   Accept ...

  6. 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 ...

  7. POJ - 1470 Closest Common Ancestors(离线Tarjan算法)

    1.输出测试用例中是最近公共祖先的节点,以及这个节点作为最近公共祖先的次数. 2.最近公共祖先,离线Tarjan算法 3. /* POJ 1470 给出一颗有向树,Q个查询 输出查询结果中每个点出现次 ...

  8. POJ 1470 Closest Common Ancestors【近期公共祖先LCA】

    版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u013912596/article/details/35311489 题目链接:http://poj ...

  9. POJ 1470 Closest Common Ancestors【LCA Tarjan】

    题目链接: http://poj.org/problem?id=1470 题意: 给定若干有向边,构成有根数,给定若干查询,求每个查询的结点的LCA出现次数. 分析: 还是很裸的tarjan的LCA. ...

随机推荐

  1. SpringBoot2.1.3修改tomcat参数支持请求特殊符号

    最近遇到一个问题,比如GET请求中,key,value中带有特殊符号,请求会报错,见如下URL: http://xxx.xxx.xxx:8081/aaa?key1=val1&a.[].id=1 ...

  2. hihocoder offer收割编程练习赛11 C 岛屿3

    思路: 并查集的应用. 实现: #include <iostream> #include <cstdio> using namespace std; ][]; int n, x ...

  3. CAS4.0 server 环境的搭建

    1.上cas的官网下载cas server 官网地址:https://github.com/Jasig/cas/releases,下载好后 解压下载的 cas-server-4.0.0-release ...

  4. mysql提升效率

    1.对查询进行优化,应尽量避免全表扫描,首先应考虑在 where 及 order by 涉及的列上建立索引. 2.应尽量避免在 where 子句中对字段进行 null 值判断,否则将导致引擎放弃使用索 ...

  5. 获取请求服务器传输协议http or https

    $protocol = (!empty($_SERVER['HTTPS']) && $_SERVER['HTTPS'] !== 'off' || $_SERVER['SERVER_PO ...

  6. 数组(Arry)几个常用方法的详解

    join() 方法用于把数组中的所有元素放入一个字符串.元素是通过指定的分隔符进行分隔的. arrayObject.join(separator)separator 可选.指定要使用的分隔符.如果省略 ...

  7. vue2.0 vue.set()

    <!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8&quo ...

  8. opencv-flag

    http://blog.csdn.net/yiyuehuan/article/details/43701797 在Mat类中定义了这样一个成员变量: /*! includes several bit- ...

  9. 禁止浏览器static files缓存篇

    由于CSS/JS文件经常需要改动,前端调试时是不希望浏览器缓存这些文件的. Meta法 目前在chrome调试还没有遇到问题,好用!此方法假设浏览器是个好人!很听我们的话! <meta http ...

  10. jq进度条

    <!doctype html><html><head><meta charset="utf-8"><title>JQue ...