HDOJ 4460 Friend Chains 图的最长路

类似于树的直径,从任意一个点出发,找到距离该点最远的且度数最少的点.

然后再做一次最短路

Friend Chains

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4227    Accepted Submission(s): 1359

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

   
   
   
   

    
    
    
    2
   
   
   
   

 

Source

2012 Asia Hangzhou Regional Contest

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月17日 星期一 16时35分00秒
File Name     :HDOJ4460.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=2100;

int n,m;
int id=1;
map<string,int> msi;


int getID(string name)
{
	if(msi[name]==0) msi[name]=id++;
	return msi[name]; 
}

struct Edge
{
	int to,next,cost;
}edge[maxn*maxn];

int Adj[maxn],Size;
int du[maxn];

void init()
{
	id=1; msi.clear();
	memset(du,0,sizeof(du));
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void Add_Edge(int u,int v)
{
	edge[Size].to=v;
	edge[Size].cost=1;
	edge[Size].next=Adj[u];
	Adj[u]=Size++;
}

int dist[maxn],cq[maxn];
bool inq[maxn];

bool spfa(int st)
{
    memset(dist,63,sizeof(dist));
    memset(cq,0,sizeof(cq));
    memset(inq,false,sizeof(inq));
    dist[st]=0;
    queue<int> q;
    inq[st]=true;q.push(st); cq[st]=1;

    while(!q.empty())
    {
        int u=q.front();q.pop();

        for(int i=Adj[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dist[v]>dist[u]+edge[i].cost)
            {
                dist[v]=dist[u]+edge[i].cost;
                if(!inq[v])
                {
                    inq[v]=true;
                    cq[v]++;
                    if(cq[v]>=n+10) return false;
                    q.push(v);
                }
            }
        }
        inq[u]=false;
    }
    return true;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d",&n)!=EOF&&n)
	{
		init();
		string name1,name2;
		for(int i=1;i<=n;i++) 
		{
			cin>>name1;
		}
		scanf("%d",&m);
		for(int i=0;i<m;i++)
		{
			cin>>name1>>name2;
			int id1=getID(name1);
			int id2=getID(name2);
			Add_Edge(id1,id2); Add_Edge(id2,id1);
			du[id1]++; du[id2]++;
		}
		spfa(1);
		int st=1;
		for(int i=2;i<=n;i++)
		{
			if(dist[st]<dist[i]) st=i;
			else if(dist[st]==dist[i])
			{
				if(du[st]>du[i]) st=i;
			}
		}
		spfa(st);
		int ans=0;
		for(int i=1;i<=n;i++) ans=max(ans,dist[i]);
		if(ans==INF) ans=-1;
		cout<<ans<<endl;
	}
    
    return 0;
}