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)
原文:http://www.cnblogs.com/dwtfukgv/p/5750576.html