UVa247
题目连接(vj,比较方便):https://vjudge.net/problem/UVA-247
Description:If you’ve seen television commercials for long-distance phone companies lately, you’ve noticed that many companies have been spending a lot of money trying to convince people that they provide the best service at the lowest cost. One company has “calling circles.” You provide a list of people that you call most frequently. If you call someone in your calling circle (who is also a customer of the same company), you get bigger discounts than if you call outside your circle. Another company points out that you only get the big discounts for people in your calling circle, and if you change who you call most frequently, it’s up to you to add them to your calling circle. LibertyBell Phone Co. is a new company that thinks they have the calling plan that can put other companies out of business. LibertyBell has calling circles, but they figure out your calling circle for you. This is how it works. LibertyBell keeps track of all phone calls. In addition to yourself, your calling circle consists of all people whom you call and who call you, either directly or indirectly. For example, if Ben calls Alexander, Alexander calls Dolly, and Dolly calls Ben, they are all within the same circle. If Dolly also calls Benedict and Benedict calls Dolly, then Benedict is in the same calling circle as Dolly, Ben, and Alexander. Finally, if Alexander calls Aaron but Aaron doesn’t call Alexander, Ben, Dolly, or Benedict, then Aaron is not in the circle. You’ve been hired by LibertyBell to write the program to determine calling circles given a log of phone calls between people.
Input
The input file will contain one or more data sets. Each data set begins with a line containing two integers, n and m. The first integer, n, represents the number of different people who are in the data set. The maximum value for n is 25. The remainder of the data set consists of m lines, each representing a phone call. Each call is represented by two names, separated by a single space. Names are first names only (unique within a data set), are case sensitive, and consist of only alphabetic characters; no name is longer than 25 letters. For example, if Ben called Dolly, it would be represented in the data file as Ben Dolly Input is terminated by values of zero (0) for n and m.
Output
For each input set, print a header line with the data set number, followed by a line for each calling circle in that data set. Each calling circle line contains the names of all the people in any order within the circle, separated by comma-space (a comma followed by a space). Output sets are separated by blank lines.
题目大意:
来自紫书推荐的经典题目,题目大意详见紫书
解题思路:
floyd算法求传递闭包,floyd算法可以求每两点之间的最短路,同时也可以求强连通分量
顺便说一下传递闭包和强连通分量,首先传递闭包定义更广泛,可以用在图论之外的地方。然后再图论中,根据传递闭包的定义,该强连通块一定是边数最少的,而强连通分量不一定是边数最少的,只要求两两可达。他们定义出发点不同。
代码如下:
- #include<bits/stdc++.h>
- #define MAX 30
- using namespace std;
- int d[MAX][MAX];
- bool vis[MAX];
- int n,m;
- void floyd()
- {
- ; k<n; k++)
- ; i<n; i++)
- ; j<n; j++)
- d[i][j]=d[i][j]||(d[i][k]&&d[k][j]);
- }
- void init()
- {
- ; i<n; i++)
- ; j<n; j++)
- d[i][j]=i==j?:;
- memset(vis,,sizeof(vis));
- }
- int main()
- {
- int cnt;
- ;
- string name1,name2;
- string namenumber[MAX];
- map <string,int> mp;
- while(~scanf("%d%d",&n,&m)&&n)
- {
- T++;
- cnt=;
- mp.clear();
- init();
- while(m--)
- {
- int u,v;
- cin>>name1>>name2;
- if(!mp.count(name1))
- mp[name1]=cnt++;
- if(!mp.count(name2))
- mp[name2]=cnt++;
- u=mp[name1];
- v=mp[name2];
- namenumber[u]=name1;
- namenumber[v]=name2;
- d[u][v]=;
- }
- floyd();
- printf("Calling circles for data set %d:\n",T);
- ; i<n; i++)
- {
- if(vis[i])
- continue;
- cout<<namenumber[i];
- ; j<n; j++)
- {
- if(vis[j])
- continue;
- if(d[j][i]&&d[i][j])
- {
- cout<<", "<<namenumber[j];
- vis[j]=;
- }
- }
- cout<<endl;
- }
- }
- }
UVa247的更多相关文章
- UVa247 Calling Circles
Time Limit: 3000MS 64bit IO Format: %lld & %llu map存人名,floyd传递闭包,DFS查询. 输出答案的逗号后面还有个空格,被坑到了2 ...
- uva247 - Calling Circles(传递闭包+DFS)
题意:两人相互打电话(直接或间接),则在一个电话圈.即a给b打电话,b给c打电话,则a给c间接打电话. 注意:1.注意标记.2.注意输出格式. #include<iostream> #in ...
- [Uva247][Tarjan求强连通分量][Calling Circles]
题目大意: 例如:A跟B打电话,B跟C打电话,C跟A打电话..D跟E打电话,E跟D不打电话.则A,B,C属于同一个电话圈,D,E分别属于一个电话圈,问有多少个电话圈. 分析 就是裸的求强连通分量,直接 ...
- 紫书 例题11-4 UVa247 (Floyd判断联通)
Floyd联通, 然后为了输出联通分量而新建一个图, 让互相可以打电话的建立一条边, 然后dfs输出联通分量就ok了. #include<cstdio> #include<iostr ...
- 洛谷 题解 UVA247 【电话圈 Calling Circles】
[题意] 如果两个人互相打电话(直接或者间接),则说他们在同一个电话圈里.例如,\(a\)打给\(b\),\(b\)打给\(c\),\(c\)打给\(d\),\(d\)打给\(a\),则这四个人在同一 ...
- UVA 247 电话圈(Floyd传递闭包+输出连通分量)
电话圈 紫书P365 [题目链接]电话圈 [题目类型]Floyd传递闭包+输出连通分量 &题解: 原来floyd还可以这么用,再配合连通分量,简直牛逼. 我发现其实求联通分量也不难,就是for ...
- UVa 247 电话圈(Floyd传递闭包)
https://vjudge.net/problem/UVA-247 题意: 如果两个人相互打电话,则说他们在同一个电话圈里.例如,a打给b,b打给c,c打给d,d打给a,则这4个人在同一个圈里:如果 ...
- 简单的floyd——初学
前言: (摘自https://www.cnblogs.com/aininot260/p/9388103.html): 在最短路问题中,如果我们面对的是稠密图(十分稠密的那种,比如说全连接图),计算多 ...
- [笔记-图论]Floyd
用于可带负权的多源最短路 时间复杂度O(n^3) 注意一定不要给Floyd一个带负环的图,不然就没有什么意义了(最短路不存在) 模板 // Floyd // to get minumum distan ...
随机推荐
- Spark程序
Spark认识&环境搭建&运行第一个Spark程序 2017-07-09 17:17 by 牛仔裤的夏天, 181 阅读, 0 评论, 收藏, 编辑 摘要:Spark作为新一代大数据计 ...
- 【bzoj1123】[POI2008]BLO DFS树
题目描述 Byteotia城市有n个 towns m条双向roads. 每条 road 连接 两个不同的 towns ,没有重复的road. 所有towns连通. 输入 输入n<=100000 ...
- springboot 实现自定义注解
1.定义一个注解@Target(ElementType.METHOD)@Retention(RetentionPolicy.RUNTIME)@Documentedpublic @interface T ...
- [Leetcode] Anagrams 颠倒字母构成词
Given an array of strings, return all groups of strings that are anagrams. Note: All inputs will be ...
- THUSC2014酱油记
Day0: 坐飞机到北京,然后报到...跟jason_yu分到一个房间,刚好可以蹭点RP.发现房间460RMB/晚,但再带一份早餐就500RMB,难道早餐是40RMB么...在一家川菜馆吃的午晚餐,感 ...
- 工具——代码中自动生成SVN版本号
本节和大家讨论一下程序集版本最后一位使用SVN版本号的自动生成方法,这里就向大家简单介绍一下.在进行自动部署的时候,经常需要用脚本获取程序的最新版本号.现在我们定义每个程序集的版本信息的最末段表示SV ...
- 修改innodb_flush_log_at_trx_commit参数提升insert性能
最近,在一个系统的慢查询日志里发现有个insert操作很慢,达到秒级,并且是比较简单的SQL语句,把语句拿出来到mysql中直接执行,速度却很快. 这种问题一般不是SQL语句本身的问题,而是在具体的应 ...
- git 的证书重新设置,以及如何让git 记住提交的用户名和密码
1.git 的证书的重新设置的命令是: git config --system --unset credential.helper 2.保存git的用户名和密码注意这里是全局保存 git config ...
- 两个数组的交集 II [ LeetCode - 350 ]
原题地址:https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/description/ 给定两个数组,写一个方法来计算 ...
- Step-By-Step: Setting up Active Directory in Windows Server 2016
There are interesting new features now made available in Windows Server 2016 such as time based grou ...