Fox and Dinner

题目描述

Fox Ciel is participating in a party in Prime Kingdom. There are $n$ foxes there (include Fox Ciel). The $i$-th fox is $a_i$ years old.

They will have dinner around some round tables. You want to distribute foxes such that:

• each fox is sitting at some table,
• each table has at least $3$ foxes sitting around it,
• the sum of ages of any two adjacent foxes around each table should be a prime number.

If $k$ foxes $f_1, f_2, \ldots, f_k$ are sitting around table in clockwise order, then for $1 \le i \le k-1$: $f_i$ and $f_{i+1}$ are adjacent, and $f_1$ and $f_k$ are also adjacent.

If it is possible to distribute the foxes in the desired manner, find out a way to do that.

题意概述

给定$n$个数$a_i$，你需要将它们分成若干个圈，使得每个圈中至少有$3$个数，且任意两个相邻的数的和为质数。

数据范围：$3 \le n \le 200, \ 2 \le a_i \le 10000$。

算法分析

因为每个数都大于$1$，所以任意两个相邻的数必定是一奇一偶。对于任意一个奇数$a_i$和任意一个偶数$a_j$，如果$a_i+a_j$是质数，那么就由$a_i$向$a_j$连一条流量为$1$的边。由于每个数都必须与另外两个数相连，因此可以由一个假设的源点向每个奇数的点连一条流量为$2$的边，由每个偶数的点向一个假设的汇点连一条流量为$2$的边。找出这张图的最大流，如果源点连出的边和连向汇点的边全部满流则说明有解。最后遍历找出所有环即可。

代码

#include <iostream>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
struct edge {
    int v, f, nxt;
} e[50001];
int n, nume = 1, top, src, sink, a[202], h[202], g[202], dist[202], que[202];
bool vis[202];
vector<int> ans[201];
void add_edge(int u, int v, int f) {
    e[++nume].v = v, e[nume].f = f, e[nume].nxt = h[u], h[u] = nume;
    e[++nume].v = u, e[nume].f = 0, e[nume].nxt = h[v], h[v] = nume;
}
void bfs() {
    memset(dist, 0, sizeof(dist));
    que[1] = src, dist[src] = 1;
    int s = 0, t = 1;
    while (s < t) {
        int u = que[++s];
        for (int i = h[u]; i; i = e[i].nxt) {
            if (e[i].f && !dist[e[i].v]) {
                que[++t] = e[i].v, dist[e[i].v] = dist[u] + 1;
            }
        }
    }
}
int dfs(int u, int delta) {
    if (u == sink) return delta;
    int ret = 0;
    for (int i = g[u]; i; i = e[i].nxt) {
        if (e[i].f && dist[e[i].v] == dist[u] + 1) {
            int dd = dfs(e[i].v, min(e[i].f, delta));
            e[i].f -= dd, e[i ^ 1].f += dd;
            if (e[i].f) g[u] = i;
            delta -= dd, ret += dd;
        }
    }
    return ret;
}
int max_flow() {
    int ret = 0;
    while (1) {
        bfs();
        if (!dist[sink]) return ret;
        for (int i = 0; i <= n + 1; ++i) g[i] = h[i];
        ret += dfs(src, 1e9);
    }
}
bool is_prime(int t) {
    int k = (int) sqrt(t);
    for (int i = 2; i <= k; ++i) if (!(t % i)) return false;
    return true;
}
void find(int u, int t) {
    vis[u] = true, ans[t].push_back(u);
    for (int i = h[u]; i; i = e[i].nxt) {
        if (e[i].v && e[i].v != n + 1 && e[i].f == !(a[u] & 1) && !vis[e[i].v]) find(e[i].v, t);
    }
}
int main()
{
    cin >> n;
    src = 0, sink = n + 1;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        if (a[i] & 1) add_edge(src, i, 2);
        else add_edge(i, sink, 2);
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (a[i] & 1 && !(a[j] & 1) && is_prime(a[i] + a[j])) add_edge(i, j, 1);
        }
    }
    max_flow();
    for (int i = h[0]; i; i = e[i].nxt) {
        if (e[i].f) {
            cout << "Impossible" << endl;
            return 0;
        }
    }
    for (int i = h[n + 1]; i; i = e[i].nxt) {
        if (e[i].f < 2) {
            cout << "Impossible" << endl;
            return 0;
        }
    }
    for (int i = 1; i <= n; ++i) if (!vis[i]) top++, find(i, top);
    cout << top << endl;
    for (int i = 1; i <= top; ++i) {
        cout << ans[i].size() << ' ';
        for (vector<int>::iterator iter = ans[i].begin(); iter != ans[i].end(); ++iter) {
            cout << *iter << ' ';
        }
        cout << endl;
    }
    return 0;
}