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题意：给定 n 个人，和关系，问你这个朋友圈里任意两者之间最短的距离是多少。
析：很明显的一个BFS,只要去找最长距离就好。如果不能全找到，就是-1.
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int dr[] = {0, 0, -1, 1};
const int dc[] = {-1, 1, 0, 0};
int n, m;
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
map<string, int> id;
vector<int> G[maxn];
int getid(const string &s){
return id[s];
}
int vis[maxn];
int vvis[maxn];
int d[maxn];
int bfs(int rt){
queue<int> q;
q.push(rt);
memset(vis, 0, sizeof(vis));
memset(d, -1, sizeof(d));
vis[rt] = vvis[rt] = 1;
int ans = rt;
d[rt] = 0;
int mmax = 0;
while(!q.empty()){
int u = q.front(); q.pop();
for(int i = 0; i < G[u].size(); ++i){
int v = G[u][i];
if(vis[v]) continue;
vis[v] = vvis[v] = 1;
d[v] = d[u] + 1;
if(mmax < d[u] + 1){
mmax = d[u] + 1;
ans = v;
}
q.push(v);
}
}
return ans;
}
int solve(int i){
int u = bfs(i);
int v = bfs(u);
return d[v];
}
int main(){
while(scanf("%d", &n) == 1 && n){
id.clear();
string s, s1, s2;
for(int i = 1; i <= n; ++i){
cin >> s;
id[s] = i;
G[i].clear();
}
scanf("%d", &m);
for(int i = 0; i < m; ++i){
cin >> s1 >> s2;
int u = getid(s1);
int v = getid(s2);
G[u].push_back(v);
G[v].push_back(u);
}
if(m + 1 < n){ printf("-1\n"); continue; }
memset(vvis, 0, sizeof(vvis));
int ans = 0;
for(int i = 1; i <= n; ++i)
if(!vvis[i]) ans = max(ans, solve(i));
for(int i = 1; i <= n; ++i)
if(d[i] == -1){ ans = -1; break; }
printf("%d\n", ans);
}
return 0;
}
HDU 4460 Friend Chains （BFS，最长路径）
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原文地址：http://www.cnblogs.com/dwtfukgv/p/5750576.html
