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HDU 4460 Friend Chains

February 8, 2018 • Read: 8515 • 算法阅读设置

题目链接:HDU 4460展开目录

题解展开目录

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。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。

代码展开目录

  • #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;
  • }
Last Modified: October 7, 2018
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