codeforces 512C (最大流)

题目链接：点击这里

题意：安排ｎ个人在圆桌吃饭，每个人有一个值，任意相邻的两个人的值的和必须为质数，一张桌子至少坐３个人。然后求出任意的方案。

每个数都大于２，要让相邻两个的和为质数，必须是一奇一偶。所以只需要每一个奇数左右两个偶数不相同就可以使得一张桌子至少３个人（３个人的桌子是不可能的，至少是４个人）。所以直接按照二分图建模，Ｓ到每一个奇数，流量为２，表示每个奇数相邻两个偶数；奇数到和他相加为质数的偶数，流量为１；偶数到Ｔ，流量为２；最后在残留网络上找方案，一个联通分量是一个方案。

#include <bits/stdc++.h>
using namespace std;
#define maxn 205
#define maxm 20005
#define type int
#define INF 1e8

int n, m;
int s, t;
struct Edge {
int from, to, next;
type cap, flow;
void get (int u, int a, int b, type c, type d) {
from = u; to = a; next = b; cap = c; flow = d;
}
}edge[maxm];
int tol;
int head[maxn];
int gap[maxn], dep[maxn], pre[maxn], cur[maxn];

bool is_prime[maxm];
void init () {
memset (is_prime, 1, sizeof is_prime);
is_prime[0] = is_prime[1] = 0;
for (int i = 2; i < maxm; i++) if (is_prime[i]) {
for (int j = i+i; j < maxm; j += i) is_prime[j] = 0;
}
tol = 0;
memset (head, -1, sizeof head);
}

void add_edge (int u, int v, type w, type rw=0) {
edge[tol].get(u, v,head[u],w,0);head[u]=tol++;
edge[tol].get(v, u,head[v],rw,0);head[v]=tol++;
}

type sap (int start, int end, int N) {
memset (gap, 0, sizeof gap);
memset (dep, 0, sizeof dep);
memcpy (cur, head, sizeof head);
int u = start;
pre[u] = -1;
gap[0] = N;
type ans = 0;
while (dep[start] < N) {
if (u == end) {
type Min = INF;
for (int i = pre[u]; i != -1; i = pre[edge[i^1].to])
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap-edge[i].flow;
for (int i = pre[u]; i != -1; i = pre[edge[i^1].to]) {
edge[i].flow += Min;
edge[i^1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for (int i = cur[u]; i != -1; i = edge[i].next) {
v = edge[i].to;
if (edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) {
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if (flag) {
u = v;
continue;
}
int Min = N;
for (int i = head[u]; i != -1; i = edge[i].next)
if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) {
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if (!gap[dep[u]]) return ans;
dep[u] = Min+1;
gap[dep[u]]++;
if (u != start) u = edge[pre[u]^1].to;
}
return ans;
}

int a[maxn], k;
vector <int> table[maxn], con[maxn];
bool vis[maxn];

void dfs (int u) {
if (vis[u]) return ;
table[k].push_back (u);
vis[u] = 1;
for (int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if (edge[i].flow && !vis[v]) {
table[k].push_back (v);
vis[v] = 1;
dfs (con[v][0] == u ? con[v][1] : con[v][0]);
break;
}
}
}

int main () {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
con[i].clear ();
}
init ();
int N;
s = 0, t = n+1, N = t+1;
int odd = 0, even = 0;
for (int i = 1; i <= n; i++) {
if (a[i]&1) {
odd++;
add_edge (s, i, 2);
for (int j = 1; j <= n; j++) if (a[j]%2 == 0 && is_prime[a[i]+a[j]]) {
add_edge (i, j, 1);
}
}
else {
even++;
add_edge (i, t, 2);
}
}
if (odd != even) {
cout << "Impossible" << endl;
return 0;
}
int ans = sap (s, t, N);
if (ans != even*2) {
cout << "Impossible" << endl;
return 0;
}
k = 0;
memset (vis, 0, sizeof vis);
for (int i = 1; i <= n; i++) if (a[i]&1) {
for (int j = head[i]; j != -1; j = edge[j].next) {
int v = edge[j].to;
if (edge[j].flow) {
con[v].push_back (i);
}
}
}
for (int i = 1; i <= n; i++) if ((a[i]&1) && !vis[i]) {
table[++k].clear ();
dfs (i);
}
cout << k << endl;
for (int i = 1; i <= k; i++) {
int sz = table[i].size ();
cout << sz << " ";
for (int j = 0; j < sz; j++) {
cout << table[i][j] << (j == sz-1 ? '\n' : ' ');
}
}
return 0;
}