Friend Chains

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 4 Accepted Submission(s) : 2

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

For a group of people, there is an idea that everyone is equals to or less than 6 steps away from any other person in the group, by way of introduction. So that a chain of "a friend of a friend" can be made to connect any 2 persons and it contains no more than 7 persons.
For example, if XXX is YYY’s friend and YYY is ZZZ’s friend, but XXX is not ZZZ's friend, then there is a friend chain of length 2 between XXX and ZZZ. The length of a friend chain is one less than the number of persons in the chain.
Note that if XXX is YYY’s friend, then YYY is XXX’s friend. Give the group of people and the friend relationship between them. You want to know the minimum value k, which for any two persons in the group, there is a friend chain connecting them and the chain's length is no more than k .

Input

There are multiple cases.
For each case, there is an integer N (2<= N <= 1000) which represents the number of people in the group.
Each of the next N lines contains a string which represents the name of one people. The string consists of alphabet letters and the length of it is no more than 10.
Then there is a number M (0<= M <= 10000) which represents the number of friend relationships in the group.
Each of the next M lines contains two names which are separated by a space ,and they are friends.
Input ends with N = 0.

Output

For each case, print the minimum value k in one line.
If the value of k is infinite, then print -1 instead.

Sample Input

3
XXX
YYY
ZZZ
2
XXX YYY
YYY ZZZ
0

Sample Output

/*
题意：给出一个无向图，若联通则输出任意一对点之间最短路径中的最大值，
若有孤立一点，则输出-1。
*/
#include <iostream>
#include<cstdio>
#include<queue>
#include<map>
#include<vector>
using namespace std;
const int inf=10000;
int n,m,maxn;
map<string,int> mp;
vector<int> s[1005];
int dis[1005];
bool vis[1005];
void spfa(int st)
{for(int i=1;i<=n;i++) dis[i]=inf,vis[i]=0;queue<int> Q;Q.push(st);vis[st]=1;dis[st]=0;while(!Q.empty()){int u=Q.front();vis[u]=0;Q.pop();for(int i=0;i<s[u].size();i++){if (dis[s[u][i]]<=dis[u]+1) continue;dis[s[u][i]]=dis[u]+1;maxn=dis[s[u][i]]>maxn?dis[s[u][i]]:maxn;if (!vis[s[u][i]]){vis[s[u][i]]=1;Q.push(s[u][i]);}}}return;
}
int main()
{while(scanf("%d",&n) && n){for(int i=1;i<=n;i++){char ch[15];scanf("%s",&ch);mp[ch]=i;s[i].clear();}scanf("%d",&m);for(int i=1;i<=m;i++){char ch1[15],ch2[15];scanf("%s %s",&ch1,&ch2);s[mp[ch1]].push_back(mp[ch2]);s[mp[ch2]].push_back(mp[ch1]);}int ans=0;for(int i=1;i<=n;i++){maxn=0;spfa(i);if (maxn==0) ans=-1;//本来想直接退出，输出-1，但是考虑到现在以i为起点的最短路径计算出来的并不一定是最长的长度，可能是最短路径中的一个中间点if (ans>=0 && maxn>ans) ans=maxn;}printf("%d\n",ans);}return 0;
}

转载于:https://www.cnblogs.com/stepping/p/5744762.html

HDU 4460 Friend Chains(map + spfa)相关推荐

hdu 4460 friend chains spfa 最短路里面的最长路
题意不再赘述... 连接http://acm.hdu.edu.cn/showproblem.php?pid=4460 此题直接郁闷致死.... 比赛的时候用的是floyd然后直接超时...当时闪过sp ...
ACM学习历程—HDU 2112 HDU Today（map spfa 优先队列）
Description 经过锦囊相助,海东集团终于度过了危机,从此,HDU的发展就一直顺风顺水,到了2050年,集团已经相当规模了,据说进入了钱江肉丝经济开发区500强.这时候,XHD夫妇也退居了二线 ...
HDU 2112 HDU Today （dijkstar + map）
HDU Today Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total ...
hdu 1263 水果 (嵌套 map)
水果 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submiss ...
最短路算法详解（Dijkstra/SPFA/Floyd）
转自:http://blog.csdn.net/murmured/article/details/19281031 一.Dijkstra Dijkstra单源最短路算法,即计算从起点出发到每个点的最短 ...
WaWa的奇妙冒险（第三周集训自闭现场）
第三周集训自闭现场(a* ida* dbfs真的好难) (一)wlacm例题记录 A-有重复元素的排列问题 (水题,但卡我到自闭) Input Output Sample Input Sample O ...
Spring Boot 整合 Shiro
虽然,直接用Spring Security和SpringBoot 进行"全家桶式"的合作是最好不过的,但现实总是欺负我们这些没办法决定架构类型的娃子. Apache Shiro 也 ...
c语言随机产生三个大写字母,C语言编写的随机产生四则运算测试题
题目:编写一个四则运算测试题的程序,要求每道题都要随机产生解题思路: 1.编写测试题,且为30道,就要用到循环函数,因此想到用for()函数 2.随机产生两个数,就想到用rand()函数. 注:1. ...
超级账本hyperledger fabric第五集：共识排序及源码阅读
一.共识机制达成共识需要3个阶段,交易背书,交易排序,交易验证交易背书:模拟的交易排序:确定交易顺序,最终将排序好的交易打包区块分发交易验证:区块存储前要进行一下交易验证二.orderer节 ...

HDU 4460 Friend Chains(map + spfa)