思路：

　　观察到路径上除了终点起点以外的每个点出度和入度都为1，和网络流的拆点很像，所以就把每个点都拆成两个点，若存在一条路径$(u,v)$，则建一条$（u,v+n,1）$的边，然后求出最大流后，每个起点的入度都是0，所以$ans=n-maxflow$。

　　注意由于是拆点，所以各种数组都要开两倍空间。

#include
     
      #define clr(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;const ll INFLL = 0x3f3f3f3f3f3f3f3f;const int INF = 0x3f3f3f3f;const int maxn = 510;struct Edge {    int to, flow, nxt;    Edge(){}    Edge(int to, int nxt, int flow):to(to),nxt(nxt), flow(flow){}}edge[maxn * maxn * 2];int head[maxn*2], dep[maxn*2];int S, T;int N, n, m, tot;void init(int n){    N=n;    for (int i = 0; i <= N; ++i) head[i] = -1;    tot = 0;}void addv(int u, int v, int w, int rw = 0){    edge[tot] = Edge(v, head[u], w); head[u] = tot++;    edge[tot] = Edge(u, head[v], rw); head[v] = tot++;}bool BFS(){    for (int i = 0; i <= N; ++i) dep[i] = -1;    queue
      
       q;    q.push(S);    dep[S] = 1;    while (!q.empty())    {        int u = q.front();        q.pop();        for (int i = head[u]; ~i; i = edge[i].nxt)        {                        if (edge[i].flow && dep[edge[i].to] == -1)            {                dep[edge[i].to] = dep[u] + 1;                q.push(edge[i].to);            }        }    }    return dep[T] < 0 ? 0 : 1;}int DFS(int u, int f){    if (u == T || f == 0) return f;    int w, used = 0;    for (int i = head[u]; ~i; i = edge[i].nxt)    {        if (edge[i].flow && dep[edge[i].to] == dep[u] + 1)        {            w = DFS(edge[i].to, min(f - used, edge[i].flow));            edge[i].flow -= w;            edge[i ^ 1].flow += w;            used += w;            if (used == f) return f;        }    }    if (!used) dep[u] = -1;    return used;}int Dicnic(){    int ans = 0;    while (BFS())    {        ans += DFS(S, INF);    }    return ans;}int main(){    cin>>n>>m;    T=2*n+1;    init(T);    S=0;    while(m--){        int u,v;        scanf("%d%d",&u,&v);        addv(u,v+n,1);    }    for(int i=1;i<=n;i++){        addv(S,i,1);        addv(i+n,T,1);    }    int ans=n-Dicnic();    printf("%d\n",ans);}