P2944 [USACO09MAR]地震损失2Earthquake Damage 2

题意：确定几个点不割，问最少割几个点使确定的点不能到达T
拆点，每个点$i$分成$i$和$i+n$

• 对于确定不能拆的点，$i$连向$i+n$的边权为$inf$，同理，其他点$i$连向$i+n$的边权为$1$
• 设置超级源点S，S连向所有确定的点，边权为$inf$，这里要注意S不能连向所有的点，因为题意是说这几个确定的点最后不能到T，而不是所有的点不能到T

#include 
using namespace std;
#define rep(i,a,n) for (int i=a;i
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

const int maxn = 1e4+100;
const int maxm = 1e5+100;
const int inf = 0x7f7f7f7f;

typedef struct Dinic
{
	typedef struct Edge
	{
		int u,v,w,nxt;	
	} Edge;
	int head[maxn],hcnt;
	int dep[maxn];
	int cur[maxn];
	Edge e[maxm];
	int S,T,N;
	void init()
	{
		memset(head,-1,sizeof head);
		hcnt = 0;
		S = T = N = 0;
	}
	void adde(int u,int v,int w)
	{
		e[hcnt].u = u,e[hcnt].v = v,e[hcnt].w = w;
		e[hcnt].nxt = head[u];head[u] = hcnt++;
		e[hcnt].u = v,e[hcnt].v = u,e[hcnt].w = 0;
		e[hcnt].nxt = head[v];head[v] = hcnt++;
	}
	int  bfs()
	{
		rep(i,0,N)
		{
			dep[i] = inf;
		}
		queue<int> q;
		q.push(S); dep[S] = 0;
		while(!q.empty())
		{
			int u = q.front();q.pop();
			for(int i = head[u];~i;i = e[i].nxt)
			{
				int v = e[i].v,w = e[i].w;
				if(w > 0 && dep[u] + 1 < dep[v])
				{
					dep[v] = dep[u] + 1;
					if(v == T)
					{
						return 1;
					}
					q.emplace(v);
				}
			}
		}
		return dep[T] != inf;
	}
	int dfs(int s,int mw)
	{
		if(s == T) return mw;
		for(int i = cur[s];~i;i=e[i].nxt)
		{
			cur[s] = i;
			int v = e[i].v,w=e[i].w;
			if(w <= 0 || dep[v] != dep[s] + 1)
			{
				continue;
			}
			int cw = dfs(v,min(w,mw));
			if(cw <= 0)
				continue;
			e[i].w -= cw;
			e[i^1].w += cw;
			return cw;				
		}
		return 0;
	}
	ll dinic()
	{
		ll res = 0;
		while(bfs())
		{
			rep(i,0,N)
			{
				cur[i] = head[i];
			}
			while(int d = dfs(S,inf))
			{
				res += 1ll * d;
			}
		}
		return res;
	}
}  Dinic;

int n,m,s;

int vis[maxn];
int main(int argc, char const *argv[])
{
	while(scanf("%d%d%d",&n,&m,&s)!=EOF)
	{
		memset(vis,0,sizeof vis);
		Dinic din; din.init();
		din.S = 0,din.T = 1,din.N = 2*n+10;
		int x,y;
		rep(i,0,m)
		{
			scanf("%d%d",&x,&y);
			din.adde(x+n,y,inf);
			din.adde(y+n,x,inf);
		}

		rep(i,0,s)
		{
			scanf("%d",&x);
			vis[x] = 1;
			din.adde(x,x+n,inf);
			din.adde(0,x,inf);
		}

		rep(i,1,n+1)
		{
			//din.adde(i+n,2*n+1,inf);
			if(vis[i] == 0)
			din.adde(i,i+n,1);
		}

		printf("%lld\n",din.dinic());

	}
	return 0;
}