题意
一共有P个牧场.由C条双向路连接.两个牧场间可能有多条路.一条路也可能连接相同的牧场.牛棚坐落在牧场1.
N (1 <= N <= P) 只奶牛打来了求救电话，说她们的农场没有被摧毁，但是已经无法到达牛棚. 求出最少可能有多少牧场被摧毁.
求最小割
没有被摧毁的连INF 注意这里的是点，要拆成两个才能转化成边（# include# define IL inline# define RG register# define Fill(a, b) memset(a, b, sizeof(a))# define Copy(a, b) memcpy(a, b, sizeof(a))using namespace std;typedef long long ll;const int _(6010), __(1e5 + 10), INF(1e8);IL ll Read(){ RG char c = getchar(); RG ll x = 0, z = 1; for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1; for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48); return x * z;}int p, c, n, fst[_], nxt[__], to[__], cnt, S, T, lev[_], max_flow, w[__], cur[_], vis[_];queue Q;IL void Add(RG int u, RG int v, RG int f){ w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++; }IL int Dfs(RG int u, RG int maxf){ if(u == T) return maxf; RG int ret = 0; for(RG int e = fst[u]; e != -1; e = nxt[e]){ if(lev[to[e]] != lev[u] + 1 || !w[e]) continue; RG int f = Dfs(to[e], min(w[e], maxf - ret)); ret += f; w[e ^ 1] += f; w[e] -= f; if(ret == maxf) break; } if(!ret) lev[u] = 0; return ret;}IL bool Bfs(){ Q.push(S); Fill(lev, 0); lev[S] = 1; while(!Q.empty()){ RG int u = Q.front(); Q.pop(); for(RG int e = fst[u]; e != -1; e = nxt[e]){ if(lev[to[e]] || !w[e]) continue; lev[to[e]] = lev[u] + 1; Q.push(to[e]); } } return lev[T];}int main(RG int argc, RG char* argv[]){ Fill(fst, -1); p = Read(); c = Read(); n = Read(); T = p + p + 1; S = 1; Add(S, S + p, INF); Add(S + p, S, 0); for(RG int i = 1, a, b; i <= c; i++) a = Read(), b = Read(), Add(a + p, b, INF), Add(b + p, a, INF); for(RG int i = 1; i <= n; i++) vis[Read()] = 1; for(RG int i = 1; i <= p; i++) if(!vis[i]) Add(i, i + p, 1), Add(i + p, i, 0); else Add(i, i + p, INF), Add(i + p, i, 0), Add(i + p, T, INF), Add(T, i + p, 0); while(Bfs()) Copy(cur, fst), max_flow += Dfs(S, INF); printf("%d\n", max_flow); return 0;}