Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF510E Fox And Dinner

LuoGu: CF510E Fox And Dinner

CF: E. Fox And Dinner

大于 $2$ 的质数一定是奇数,两个数的和是奇数,两个数一定是一奇一偶,考虑建出二分图,做最大流,然后 dfs 找答案。

查看代码
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#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...rest>
void read(Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write(Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
const int N = 210, M = 1e5 + 10, inf = 200;
namespace PrimeSeive
{
const int N = 2e4 + 10;
bool vis[N];
int cnt, p[N];
void init()
{
for (int i = 2; i < N; i++)
{
if (!vis[i]) p[++cnt] = i;
for (int j = 1; j <= cnt && i * p[j] < N; j++)
{
vis[i * p[j]] = true;
if (i % p[j] == 0)
break;
}
}
}
}
bool vis[N];
int n, st, ed, w[N], d[N], cur[N];
int idx = -1, hd[N], nxt[M], edg[M], wt[M];
void add (int a, int b, int c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
bool bfs()
{
for (int i = st; i <= ed; i++)
d[i] = -1;
d[st] = 0;
cur[st] = hd[st];
queue <int> q;
q.push(st);
while (!q.empty())
{
int t = q.front();
q.pop();
for (int i = hd[t]; ~i; i = nxt[i])
if (d[edg[i]] == -1 && wt[i])
{
cur[edg[i]] = hd[edg[i]];
d[edg[i]] = d[t] + 1;
if (edg[i] == ed)
return true;
q.push(edg[i]);
}
}
return false;
}
int exploit(int x, int limit)
{
if (x == ed)
return limit;
int res = 0;
for (int i = cur[x]; ~i && res < limit; i = nxt[i])
{
cur[x] = i;
if (d[edg[i]] == d[x] + 1 && wt[i])
{
int t = exploit(edg[i], min(wt[i], limit - res));
if (!t)
d[edg[i]] = -1;
wt[i] -= t;
wt[i ^ 1] += t;
res += t;
}
}
return res;
}
int dinic()
{
int res = 0, flow;
while (bfs())
while (flow = exploit(st, inf))
res += flow;
return res;
}
void dfs (int x, vector <int> &p)
{
vis[x] = true;
p.push_back(x);
for (int i = hd[x]; ~i; i = nxt[i])
if (!vis[edg[i]] && wt[i ^ (w[x] & 1)])
return dfs(edg[i], p);
}
int main ()
{
PrimeSeive::init();
read(n);
st = 0, ed = n + 1;
for (int i = st; i <= ed; ++i) hd[i] = -1;
vector <int> odd, even;
for (int i = 1; i <= n; i++)
{
read(w[i]);
if (w[i] & 1)
{
add(st, i, 2), add(i, st, 0);
odd.push_back(i);
}
else
{
add(i, ed, 2), add(ed, i, 0);
even.push_back(i);
}
}
for (int i : odd) for (int j : even)
if (!PrimeSeive::vis[w[i] + w[j]])
add(i, j, 1), add(j, i, 0);
if (dinic() ^ n) return puts("Impossible"), 0;
vector <vector <int>> ans;
for (int i : odd) if (!vis[i])
{
vector <int> p;
dfs(i, p);
ans.push_back(p);
}
write(ans.size()), puts("");
for (auto i : ans)
{
write(i.size()), putchar(' ');
for (int j : i)
write(j), putchar(' ');
puts("");
}
return 0;
}