题目链接: hdu4460 ( Friend Chains )

将每个人当作顶点，朋友关系当作顶点之间权值为 \(1\) 的边，建立无向图。

图上任意两点间最短距离的最大值即为所求，从每个顶点 \(bfs\) 就最大距离即可，时间复杂度 \(O(n^2)\) 。

特别地，若该图不连通，则输出 \(-1\) 。

/**
* hdu4460 Friend Chains
*
*/

#include <cstdio>
#include <queue>
#include <iostream>
#include <cstring>
#include <string>
#include <stack>
#include <vector>
#include <map>
using namespace std;

const int N = 1005;

vector<int> G[N];
bool vis[N];
int n, m;

struct  node{int x, step;};
int bfs(int s)
{
    memset(vis, 0, sizeof vis);
    queue<node>Q;
    Q.push(node{s, 0});
    vis[s] = 1;
    int ans = 0;
    while (!Q.empty()) {
        node t = Q.front(); Q.pop();
        ans = max(ans, t.step);
        for (int i = 0; i < G[t.x].size(); ++i) {
            int y = G[t.x][i];
            if (vis[y]) continue;
            Q.push(node{y, t.step+1});
            vis[y] = 1;
        }
    }
    return ans;
}
int main()
{
    while (cin >> n && n) {
        string s;
        int cnt = 0;
        map<string, int> mp;
        for (int i = 1; i <= n; ++i) cin >> s, mp[s] = ++cnt;
        cin >> m;

        memset(vis, 1, sizeof vis);
        for (int i = 1; i <= n; ++i) G[i].clear();
        for (int i = 0; i < m; ++i) {
            string a, b;
            cin >> a >> b;
            int x = mp[a];
            int y = mp[b];
            G[x].push_back(y);
            G[y].push_back(x);
            vis[x] = vis[y] = 0;
        }

        bool flag = false;
        for (int i = 1; i <= n; ++i) {
            if (vis[i]) {
                cout << "-1" << endl;
                flag = true;
                break;
            }
        }
        if (flag) continue;

        int ans = 0;
        for (int i = 1; i <= n; ++i) ans = max(ans, bfs(i));
        cout << ans << endl;
    }
    return 0;
}