CF510E Fox And Dinner

题面

题解

首先可以发现：由于(a_i geq 2)，所以质数肯定是被拆成一个奇数和一个偶数。

这样的话很类似一个二分图模型，所以考虑网络流。

当(a_i)是奇数时连边((S, i, 2))，当(a_i)是偶数时连边((i, T, 2))，表示一个点的邻居最多有两个点。

若(a_i)是奇数，(a_j)是偶数，(a_i + a_j)是质数，则连边((i, j, 1))。

跑出来如果最大流不是(n)则不合法，否则直接暴力找环即可。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

inline int read()
{
	int data = 0, w = 1; char ch = getchar();
	while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if (ch == '-') w = -1, ch = getchar();
	while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data * w;
}

const int N(205), LIM(20010), M(20000), INF(0x3f3f3f3f);
struct edge { int next, to, cap; } e[N * N * 2];
int n, A[N], not_prime[LIM], head[N], e_num = -1, lev[N], cur[N], S, T;
inline void Add(int from, int to, int cap)
{
	e[++e_num] = (edge) {head[from], to, cap}, head[from] = e_num;
	e[++e_num] = (edge) {head[to], from,  0 }, head[to]   = e_num;
}

void Init()
{
	not_prime[1] = 1;
	for (int i = 2; i <= M; i++) if (!not_prime[i])
		for (int j = i * i; j <= M; j += i) not_prime[j] = 1;
}

int bfs()
{
	std::queue<int> Q; Q.push(S);
	memset(lev, 0, (T + 1) << 2), lev[S] = 1;
	while (!Q.empty())
	{
		int x = Q.front(); Q.pop();
		for (int i = head[x]; ~i; i = e[i].next)
		{
			int to = e[i].to; if (!e[i].cap || lev[to]) continue;
			lev[to] = lev[x] + 1, Q.push(to);
		}
	}
	return lev[T];
}

int dfs(int x, int f)
{
	if (x == T || !f) return f;
	int ans = 0, cap;
	for (int i = head[x]; ~i; i = e[i].next)
	{
		int to = e[i].to;
		if (e[i].cap && lev[to] == lev[x] + 1)
		{
			cap = dfs(e[i].to, std::min(f - ans, e[i].cap));
			e[i].cap -= cap, e[i ^ 1].cap += cap, ans += cap;
			if (ans == f) break;
		}
	}
	if (!ans) lev[x] = 0;
	return ans;
}

int Dinic()
{
	int ans = 0;
	while (bfs()) memcpy(cur, head, (T + 1) << 2), ans += dfs(S, INF);
	return ans;
}

std::vector<int> cir[N]; int tot, vis[N];
void find(int x)
{
	if (1 <= x && x <= n) cir[tot].push_back(x); vis[x] = 1;
	for (int i = head[x]; ~i; i = e[i].next)
		if (!vis[e[i].to] && e[i | 1].cap == 1) find(e[i].to);
}

int main()
{
	memset(head, -1, sizeof head);
	Init(), n = read(), S = 0, T = n + 1; int p = 0, q = 0;
	for (int i = 1; i <= n; i++) A[i] = read(), ((A[i] & 1) ? ++p : ++q);
	if (p != q) return puts("Impossible"), 0;
	for (int i = 1; i <= n; i++)
		if (A[i] & 1) for (int j = (Add(S, i, 2), 1); j <= n; j++)
			{ if (!not_prime[A[i] + A[j]]) Add(i, j, 1); }
		else Add(i, T, 2);
	if (Dinic() != n) return puts("Impossible"), 0;
	for (int i = 1; i <= n; i++) if (!vis[i]) ++tot, find(i);
	printf("%d
", tot);
	for (int i = 1; i <= tot; i++, puts(""))
	{
		printf("%lu ", cir[i].size());
		for (int j : cir[i]) printf("%d ", j);
	}
	return 0;
}