Bus Pass


Time Limit: 5 Seconds      Memory Limit: 32768 KB

You travel a lot by bus and the costs of all the seperate tickets are starting to add up.

Therefore you want to see if it might be advantageous for you to buy a bus pass.

The way the bus system works in your country (and also in the Netherlands) is as follows:

when you buy a bus pass, you have to indicate a center zone and a star value. You are allowed to travel freely in any zone which has a distance to your center zone which is less than your star value. For example, if you have a star value of one, you can only travel in your center zone. If you have a star value of two, you can also travel in all adjacent zones, et cetera.

You have a list of all bus trips you frequently make, and would like to determine the minimum star value you need to make all these trips using your buss pass. But this is not always an easy task. For example look at the following figure:

Here you want to be able to travel from A to B and from B to D. The best center zone is 7400, for which you only need a star value of 4. Note that you do not even visit this zone on your trips!

Input

On the first line an integert(1 <=t<= 100): the number of test cases. Then for each test case:

One line with two integersnz(2 <=nz<= 9 999) andnr(1 <=nr<= 10): the number of zones and the number of bus trips, respectively.

nz lines starting with two integers idi (1 <= idi <= 9 999) and mzi (1 <= mzi <= 10), a number identifying the i-th zone and the number of zones adjacent to it, followed by mzi integers: the numbers of the adjacent zones.

nr lines starting with one integer mri (1 <= mri <= 20), indicating the number of zones the ith bus trip visits, followed by mri integers: the numbers of the zones through which the bus passes in the order in which they are visited.

All zones are connected, either directly or via other zones.

Output

For each test case:

One line with two integers, the minimum star value and the id of a center zone which achieves this minimum star value. If there are multiple possibilities, choose the zone with the lowest number.

Sample Input

1
17 2
7400 6 7401 7402 7403 7404 7405 7406
7401 6 7412 7402 7400 7406 7410 7411
7402 5 7412 7403 7400 7401 7411
7403 6 7413 7414 7404 7400 7402 7412
7404 5 7403 7414 7415 7405 7400
7405 6 7404 7415 7407 7408 7406 7400
7406 7 7400 7405 7407 7408 7409 7410 7401
7407 4 7408 7406 7405 7415
7408 4 7409 7406 7405 7407
7409 3 7410 7406 7408
7410 4 7411 7401 7406 7409
7411 5 7416 7412 7402 7401 7410
7412 6 7416 7411 7401 7402 7403 7413
7413 3 7412 7403 7414
7414 3 7413 7403 7404
7415 3 7404 7405 7407
7416 2 7411 7412
5 7409 7408 7407 7405 7415
6 7415 7404 7414 7413 7412 7416

Sample Output

4 7400

参考:http://blog.csdn.net/acvay/article/details/40221819

题意:从给的公交线路(nr)找到一个中心,这个中心到每一个公交车站距离最优即(这个中心顾及到每个公交车站)。并求出这个中心到其中一个公交车站最远距离。

思路:对每一个公交车站进行BFS求出dis[i],i这个点到这些公交车站最远距离。

收获:1.了解了set:每个元素最多只出现一次,并且默认的是从小到大排序。

2.set<int>::iterator 从set中将编号取出来。

 #include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
#define maxn 10000
typedef set<int>::iterator it;
set<int>zone;
int link[maxn][];
int dis[maxn],tdis[maxn];
void bfs(int st)
{
queue<int>q;
memset(tdis,,sizeof(tdis));
tdis[st]=;
q.push(st);
while(!q.empty())
{
int cur=q.front();
q.pop();
if(tdis[cur]>dis[cur])
dis[cur]=tdis[cur];
for(int i=;i<=link[cur][];i++)
{
int temp=link[cur][i];
if(tdis[temp]==)
{
q.push(temp);
tdis[temp]=tdis[cur]+;
}
}
} }
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
zone.clear();
int nz,nr,mz,mr,num,id;
scanf("%d%d",&nz,&nr);
for(int i=;i<nz;i++)
{
scanf("%d%d",&id,&mz);
link[id][]=mz;
for(int j=;j<=mz;j++)
{
scanf("%d",&link[id][j]);
zone.insert(link[id][j]);
}
}
int pos;
memset(dis,,sizeof(tdis));
for(int i=;i<nr;i++)
{
scanf("%d",&mr);
for(int j=;j<mr;j++)
{
scanf("%d",&pos);
bfs(pos);
}
}
it i=zone.begin();
int ans= *i;
for(++i;i!=zone.end();++i)
if(dis[*i]<dis[ans])
ans=*i;
printf("%d %d\n",dis[ans],ans);
}
return ;
}

ZOJ2913Bus Pass(BFS+set)的更多相关文章

  1. ACM/ICPC 之 BFS范例(ZOJ2913-ZOJ1136(POJ1465))

    通过几道经典BFS例题阐述BFS思路 ZOJ2913-Bus Pass 题意:找一个center区域,使得center到所有公交线路最短,有等距的center则输出id最小的. 题解:经典的BFS,由 ...

  2. CQBZOJ 避开怪兽

    题目描述 给出一个N行M列的地图,地图形成一个有N*M个格子的矩阵.地图中的空地用'.'表示.其中某些格子有怪兽,用'+'表示.某人要从起点格子'V'走到终点格子'J',他可以向上.下.左.右四个方向 ...

  3. ZOJ 2913 Bus Pass (近期的最远BFS HDU2377)

    题意  在全部城市中找一个中心满足这个中心到全部公交网站距离的最大值最小 输出最小距离和满足最小距离编号最小的中心 最基础的BFS  对每一个公交网站BFS  dis[i]表示编号为i的点到全部公交网 ...

  4. (BFS)hdoj2377-Bus Pass

    题目地址 因为最后要看的是到所有路线上的区域最大距离最小的中心点,所以可以采取遍历路线上所有的区域,对每个区域进行BFS的办法.为了更方便的在每一次BFS都遍历所有的区域,可以加一个reach数组,记 ...

  5. 【BZOJ-1656】The Grove 树木 BFS + 射线法

    1656: [Usaco2006 Jan] The Grove 树木 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 186  Solved: 118[Su ...

  6. hiho #1372:平方求 (bfs)

    #1372 : 平方求和 时间限制:1000ms 单点时限:1000ms 内存限制:256MB 描述 对于一个非负整数n,最少需要几个完全平方数,使其和为n? 输入 输入包含多组数据.对于每组数据: ...

  7. POJ 2697 A Board Game(Trie判重+BFS)

    A Board Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 551   Accepted: 373 Descri ...

  8. HDU 5025:Saving Tang Monk(BFS + 状压)

    http://acm.hdu.edu.cn/showproblem.php?pid=5025 Saving Tang Monk Problem Description   <Journey to ...

  9. hdu 5025 Saving Tang Monk(bfs+状态压缩)

    Description <Journey to the West>(also <Monkey>) is one of the Four Great Classical Nove ...

随机推荐

  1. Inorder Successor in BST 解答

    Question Given a binary search tree and a node in it, find the in-order successor of that node in th ...

  2. 2016"百度之星" - 资格赛(Astar Round1) 1004

    思路:题目很简单,直接用map记录每个字符串的个数就可以了.记得对每个字符串先sort(). AC代码: #include <cstdio> #include <stdlib.h&g ...

  3. linux文件权限整理

    网上对linux文件权限的已经很多,不过还是要自己整理一下,不然每次都要查资料. linux下所有东西都是文件,包括设备,所以这里的文件也包括文件夹. 先是查看文件权限:ls -lh xzc@xzc- ...

  4. PC--CSS常识

    1.不要使用过小的图片做背景平铺.这就是为何很多人都不用 1px 的原因,这才知晓.宽高 1px 的图片平铺出一个宽高 200px 的区域,需要 200*200=40, 000 次,占用资源.2.无 ...

  5. Python模块学习笔记— —time与datatime

    Python提供了多个内置模块用于操作日期时间.像calendar,time,datetime.首先对time模块中最经常使用的几个函数作一个介绍,它提供的接口与C标准库time.h基本一致.然后再介 ...

  6. Linux 下提高make的编译效率

    Linux下安装程序,一般都通过包管理器安装,但是包管理器或软件商店里的软件往往不是最新版本的,安装最新版软件时通常是下载源代码进行编译. 编译安装源代码时就离不开make了,但是make是单线程的, ...

  7. PHP学习笔记二十八【抽象类】

    <?php //定义一个抽象类.主要用来被继承 //如果一个类继承了抽象类,则它必须实现该抽象类的所有抽象方法(除非它自己也是抽象类) // abstract class Animal{ pub ...

  8. add monitor resolution

    cvt 1920 1080 # Get the correct settings for the new mode # Output for me: "1920x1080_60.00&quo ...

  9. jquery file upload 后台收到的文件名中文乱码, filename中文乱码

    在jQuery File Upload.js文件里,在以下这个js中有个成员叫做 _initXHRData, 是一个function, 在这个function的最后部分有一个if-else分支,如下:

  10. php正则验证sql方注入

    <?php function inject_check($Sql_Str) {//自动过滤Sql的注入语句. $check=preg_match('/select|insert|update|d ...