[CF580C]Shortest Cycle(图论，最小环)

Description:

给 \(n\) 个点的图，点有点权 \(a_i\) ，两点之间有边当且仅当 \(a_i\ \text{and}\ a_j \not= 0\)，边权为1，求最小环。

Solution:

按每一位考虑若当前这一位为 1 的点超过了 2 个，那么答案就为 3 。

否则只会连一条边，于是最多只有 \(60\) 条边，枚举每条边删掉，求最短路 (边权为1，bfs) 即可。

#include <iostream>
#include <set>
#include <queue>
#include <cstring>
#include <cstdio>
#include <fstream>

typedef long long LL;
typedef unsigned long long uLL;

#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define MP(x, y) std::make_pair(x, y)
#define DE(x) cerr << x << endl;
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define GO cerr << "GO" << endl;

using namespace std;

inline void proc_status()
{
	ifstream t("/proc/self/status");
	cerr << string(istreambuf_iterator<char>(t), istreambuf_iterator<char>()) << endl;
}
template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }

const int maxN = 1e5 + 2;

int n;
LL a[maxN];
int dis[maxN];
bool vis[maxN];
int ans(0x3f3f3f3f);
vector<int> g[maxN];
set<pair<int, int> > S;

void add(int u, int v)
{
	g[u].push_back(v);
	g[v].push_back(u);
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("xhc.in", "r", stdin);
	freopen("xhc.out", "w", stdout);
#endif
	ios::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; ++i)
		cin >> a[i];
	for (int i = 0; i < 62; ++i)
	{
		int cnt = 0;
		for (int j = 1; j <= n; ++j)
		{
			if (a[j] >> i & 1)
				cnt++;
		}
		if (cnt >= 3)
		{
			cout << 3 << endl;
			return 0;
		}

		if (cnt != 2)
			continue;

		int first = 0, second = 0;
		for (int j = 1; j <= n; ++j)
			if (a[j] >> i & 1)
			{
				if (!first)
				{
					first = j;
				}
				else
				{
					second = j;
					break;
				}
			}
		S.insert(MP(first, second));
	}
	for (auto p : S)
		add(p.first, p.second);
	for (auto p : S)
	{
		int s = p.first, t = p.second;

		memset(vis, 0, sizeof vis);
		memset(dis, 0x3f, sizeof dis);

		vis[s] = 1;
		queue<int> q;
		q.push(s);
		dis[s] = 0;

		while (q.size())
		{
			int u = q.front();
			q.pop();
			for (int v : g[u])
			{
				if (u == s and v == t)
					continue;
				if (!vis[v])
				{
					q.push(v);
					vis[v] = 1;
					dis[v] = dis[u] + 1;
				}
			}
		}
		if (dis[t] < 0x3f3f3f3f) chkmin(ans, dis[t] + 1);
	}
	if (ans < 0x3f3f3f3f) cout << ans << endl;
	else cout << -1 << endl;
	return 0;
}