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

 

Sample Output

 

Source

2012 Asia Hangzhou Regional Contest

#include <stdio.h>

#include <map>

#include <string>

#include <queue>

#include <iostream>

#define inf 0x3f3f3f3f

#define MAXN 1005

using namespace std;

int cnt;

map< string,int > M;

int head[MAXN];

bool visited[MAXN];

int dist[MAXN][MAXN];

struct EdgeNode{

    int to;

    int next;

}edges[MAXN*];

void addedge(int u, int v){

    edges[cnt].to=v;

    edges[cnt].next=head[u];

    head[u]=cnt++;

}

void bfs(int u){

    queue<int> Q;

    Q.push(u);

    dist[u][u]=;

    memset(visited,,sizeof(visited));

    visited[u]=;

    while( !Q.empty() ){

        int now=Q.front();

        Q.pop();

        for(int i=head[now]; i!=-; i=edges[i].next){

            int to=edges[i].to;

            if(!visited[to]){

                dist[u][to]=dist[u][now]+;

                Q.push(to);

                visited[to]=;

            }

        }

    }

}

int main(int argc, char *argv[])

{

    int n,k;

    string a,b;

    while(scanf("%d",&n)!=EOF && n){

        M.clear();

        memset(head,-,sizeof(head));

        for(int i=; i<=n; i++){

            for(int j=i+; j<=n; j++){

                dist[i][j]=dist[j][i]=inf;

            }

        }

        for(int i=; i<=n; i++){

            string name;

            cin>>name;

            M[name]=i;

        }

        cnt=;

        scanf("%d",&k);

        while(k--){

            cin>>a>>b;

            addedge(M[a],M[b]);

            addedge(M[b],M[a]);

        }

        for(int i=; i<=n; i++)bfs(i);

        int ans=;

        for(int i=; i<=n; i++){

            for(int j=i+; j<=n; j++){

                if(dist[i][j]>ans)

                    ans=dist[i][j];

            }

        }

        if(ans==inf)

            printf("-1\n");

        else

           printf("%d\n",ans);

    }

    return ;

}