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的点到全部公交网站距离的最大值 bfs全然部网站后 dis[i]最小的点就是要求的点咯
#include<cstdio>
#include<cstring>
#include<queue>
#include<set>
using namespace std; typedef set<int>::iterator it;
const int N = 10000;
int dis[N], tdis[N], link[N][12];
queue<int> q;
set<int> zone; void bfs(int o)
{
memset(tdis, 0, sizeof(tdis));
tdis[o] = 1;
q.push(o);
while(!q.empty())
{
int cur = q.front();
q.pop();
if(tdis[cur] > dis[cur]) dis[cur] = tdis[cur];
for(int i = 1; i <= link[cur][0]; ++i)
{
int j = link[cur][i];
if(tdis[j] == 0) q.push(j), tdis[j] = tdis[cur] + 1;
}
}
} int main()
{
int cas, nz, nr, id, mz, mr, ans, t;
scanf("%d", &cas);
while(cas--)
{
zone.clear();
memset(dis, 0, sizeof(dis));
scanf("%d%d", &nz, &nr);
for(int i = 1; i <= nz; ++i)
{
scanf("%d %d", &id, &mz);
link[id][0] = mz;
zone.insert(id);
for(int i = 1; i <= mz; ++i)
scanf("%d", &link[id][i]);
} for(int i = 1; i <= nr; ++i)
{
scanf("%d", &mr);
for(int j = 1; j <= mr; ++j)
{
scanf("%d", &t);
bfs(t);
}
} it i = zone.begin();
ans = *i;
for(++i; i != zone.end(); ++i)
if(dis[*i] < dis[ans]) ans = *i;
printf("%d %d\n", dis[ans], ans);
}
return 0;
}
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
ZOJ 2913 Bus Pass (近期的最远BFS HDU2377)的更多相关文章
- 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> ...
- Bus Pass
ZOJ Problem Set - 2913 Bus Pass Time Limit: 5 Seconds Memory Limit: 32768 KB You travel a lot b ...
- hdu 2377 Bus Pass
Bus Pass Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- 【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 ...
- 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 ...
- 致远A8任意文件写入漏洞_getshell_exp
近期爆出致远 OA 系统的一些版本存在任意文件写入漏洞,远程攻击者在无需登录的情况下可通过向 URL /seeyon/htmlofficeservlet POST 精心构造的数据即可向目标服务器写入任 ...
- ZOJ 1091 Knight Moves(BFS)
Knight Moves A friend of you is doing research on the Traveling Knight Problem (TKP) where you are t ...
- ZOJ 1005 Jugs(BFS)
Jugs In the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with th ...
随机推荐
- [Oracle] Oracle终极解锁
一些ORACLE中的进程被杀掉后,状态被置为"killed",但是锁定的资源很长时间不释放,有时实在没办法,只好重启数据库.现在提供一种方法解决这种问题,那就是在ORACLE中杀不 ...
- 在树莓派下对多个串口转USB设备进行设备名称绑定操作
在开发过程中,需要用一个树莓派链接多个串口转USB设备(GPS模块,数传模块等),在树莓派linux系统环境下,USB串口设备的命名规则是 /dev/ttyUSB0 ,/dev/ttyUSB1,/de ...
- C - Puzzles
Problem description The end of the school year is near and Ms. Manana, the teacher, will soon have t ...
- testNG中方法的调用顺序
今天在执行selnium的test case时,总是遇到空指针错误.但是以前也有run成功过,然后换了各种方法定位元素,都失败了,所以怀疑应该不是元素定位不到的问题,所以可能是method之间有依赖, ...
- MapReduce架构与生命周期
MapReduce架构与生命周期 概述:MapReduce是hadoop的核心组件之一,可以通过MapReduce很容易在hadoop平台上进行分布式的计算编程.本文组织结果如下:首先对MapRedu ...
- docker批量删除容器、镜像
1.删除所有容器 docker rm `docker ps -a -q` docker rm $(docker ps -aq) 2.删除所有镜像 docker rmi `docker images - ...
- 推荐一款能支持国密SM2浏览器——密信浏览器
密信浏览器( MeSince Browser )是基于Chromium开源项目开发的国密安全浏览器,支持国密算法和国密SSL证书,同时也支持国际算法及全球信任SSL证书:密信浏览器使用界面清新,干净. ...
- redis 安装成功后外部服务器链接不上
1.reids服务器的6379端口telnet不通 2. 查看reids进程和端口,都是存在的.只是ip地址是127.0.0.1而不是0.0.0.0,只是本机能使用 3.查找redis的配置文件red ...
- jquery为元素绑定事件
语法 $(selector).live(event,data,function) 参数event 必需,规定附加到元素的一个或多个事件.由空格分隔多个事件,必须是有效的事件.data 可选,规定传递到 ...
- Javascript中的原型链,__proto__和prototype等问题总结
1.js中除了原始数据类型 都是对象. 包括函数也是对象,可能类似于C++函数对象把 应该是通过解释器 进行()操作符重载或其他操作, 用的时候把它当函数用就行 但是实际上本质是一个对象 原型也是一个 ...