Bus Pass
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
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
using namespace std;
const int INF=0x5fffffff;
const int MS=;
const double EXP=1e-; int nz,nr;
int num[MS];
int edge[MS][];
int res[MS];
int cur;
int reach[MS];// 防止在同一站重复访问。每一站对应一个cur,标记作用
int vis[MS];
void bfs(int s)
{
int i,a,b;
int val,at;
queue<int> que[]; //一层一层访问,可以用滚动队列
a=;b=;val=;
if(reach[s]<cur)
{
que[b].push(s);
reach[s]=cur;
res[s]=max(res[s],val);
}
while(!que[b].empty())
{
swap(a,b);
val++; //层次加1
while(!que[a].empty())
{
at=que[a].front();
que[a].pop();
for(i=;i<num[at];i++)
{
if(reach[edge[at][i]]<cur)
{
reach[edge[at][i]]=cur;
que[b].push(edge[at][i]);
res[edge[at][i]]=max(res[edge[at][i]],val);
}
}
}
}
} int main()
{
int T;
int i,j;
int id;
int mr;
int ret,center;
scanf("%d",&T);
while(T--)
{
memset(reach,,sizeof(reach));
memset(res,,sizeof(res));
memset(vis,,sizeof(vis));
cur=;
scanf("%d %d",&nz,&nr);
for(i=;i<nz;i++)
{
scanf("%d",&id);
scanf("%d",&num[id]);
for(j=;j<num[id];j++)
{
scanf("%d",&edge[id][j]);
}
}
for(i=;i<nr;i++)
{
scanf("%d",&mr);
for(j=;j<mr;j++)
{
scanf("%d",&id);
if(vis[id]==) //可能多条线路中有公共站点
{
vis[id]=;
bfs(id);
cur++;
}
}
}
ret=INF;center=-;
for(i=;i<;i++)
{
if(reach[i]==cur-&&res[i]<ret)
{
ret=res[i];
center=i;
}
}
printf("%d %d\n",ret,center);
}
return ;
}
Bus Pass的更多相关文章
- hdu 2377 Bus Pass
Bus Pass Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- 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 ...
- zoj 2913 Bus Pass
对于每个输入的站点求出所有点到这个站点的最短路.用anss数组存下来,然后就可以用anss数组求出答案了. 题目分析清楚了 还是比较水的,折腾了一早上.. #include<stdio.h> ...
- ZOJ2913Bus Pass(BFS+set)
Bus Pass Time Limit: 5 Seconds Memory Limit: 32768 KB You travel a lot by bus and the costs of ...
- hdu2377Bus Pass(构建更复杂的图+spfa)
主题链接: 啊哈哈,点我点我 思路: 题目是给了非常多个车站.然后要你找到一个社区距离这些车站的最大值最小..所以对每一个车站做一次spfa.那么就得到了到每一个社区的最大值,最后对每一个社区扫描一次 ...
- 【转】最短路&差分约束题集
转自:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★254 ...
- 【转载】图论 500题——主要为hdu/poj/zoj
转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并 ...
- HOJ题目分类
各种杂题,水题,模拟,包括简单数论. 1001 A+B 1002 A+B+C 1009 Fat Cat 1010 The Angle 1011 Unix ls 1012 Decoding Task 1 ...
- Drools文档(六) 用户手册
用户手册 基础 无状态的知识Session Drools规则引擎拥有大量的用例和功能,我们要如何开始?你无须担心,这些复杂性是分层的,你可以用简单的用例来逐步入门. 无状态Session,无须使用推理 ...
随机推荐
- [翻译]Behavior-Driven Development (BDD)行为驱动开发(一)
简单而言,BDD是一系列基于TDD的工具和方法集发展而来的开发模式,一般不认为是一种新的开发模式,而是作为TDD的补充.因此,首先对TDD的概念进行进行. 测试驱动开发(TDD) TDD模式采取的是迭 ...
- 判定元素正在插入到DOM树——DOMNodeInsertedIntoDocument
在firefox, webkit中我们可以使用DOMNodeInsertedIntoDocument事件,但这个事件很快变废弃了,虽然浏览器还是很有节操地支持它们,但哪一天不在也很难说.比如说fire ...
- quartz 的job中获取到applicationContext
第一步: 定义SchedulerFactoryBean的applicationContextSchedulerContextKey <bean name="scheduler" ...
- Intellij IDEA 杂记
添加JUnit File > Settings > Plugins > Browse repositories > 搜索junit ,安装JunitGenerator V2 重 ...
- GWT+CodeTemplate+TableCreate快速开发
刚进一家新公司,公司表示让我们几个新人写页面联系熟悉 怎么快速开发,进入正题: 1.根据设计规范设计页面excel 2.CodeTemplate根据excel生成属性类和对应方法(文本框,下拉框等等单 ...
- oracle学习 七 拼接变量+日期函数(持续更)
select count(KEYCODE) from STHSGDOC.ZJSJJL where ysrq=to_date(to_char(sysdate,'yyyy')||'/1','yyyy/MM ...
- Codeforces 444 C. DZY Loves Colors (线段树+剪枝)
题目链接:http://codeforces.com/contest/444/problem/C 给定一个长度为n的序列,初始时ai=i,vali=0(1≤i≤n).有两种操作: 将区间[L,R]的值 ...
- UVaLive 6802 Turtle Graphics (水题,模拟)
题意:给定一个坐标,和一行命令,按照命令走,问你有多少点会被访问超过一次. 析:很简单么,按命令模拟就好,注意有的点可能走了多次,只能记作一次. 代码如下: #pragma comment(linke ...
- chrome emulator use-agent 设置 chrom模拟手机客户端
谷歌升级以后,发现找不到use-agent设置了 在Element 下点击ESC 出现console,再点击Emulation就出现了
- MyArrayList——实现自己的ArrayList!
注:转载请注明原文地址:http://www.cnblogs.com/ygj0930/p/5965205.html ArrayList是我们常用的集合类之一,其实它的实现机制很简单,底层还是使用了一个 ...