裸的最小割，拆点时要考虑清楚到底是怎么连
如果之前i->i+n，之后又x->y+n，显然出不了解
所以可以改为i+n->i

如果要输出方案

考虑每一个一个点，如果删去这个点，最小割变小了，那么这个点就是所求点。

不输出方案的代码

# include <bits/stdc++.h>

# 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 _(210), __(1e5 + 10), INF(2147483647);

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 n, m, w[__], fst[_], nxt[__], to[__], cnt;

int S, T, lev[_], cur[_], max_flow;

queue <int> 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++;

    w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;

}

IL int Dfs(RG int u, RG int maxf){

    if(u == T) return maxf;

    RG int ret = 0;

    for(RG int &e = cur[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;

    }

    return ret;

}

IL bool Bfs(){

    Fill(lev, 0); lev[S] = 1; Q.push(S);

    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[]){

    n = Read(); m = Read(); S = Read(); T = Read() + n; Fill(fst, -1);

    for(RG int i = 1; i <= n; i++) Add(i + n, i, 1);

    for(RG int i = 1, a, b; i <= m; i++) a = Read(), b = Read(), Add(a, b + n, 1), Add(b, a + n, 1);

    while(Bfs()) Copy(cur, fst), max_flow += Dfs(S, INF);

    printf("%d\n", max_flow);

    return 0;

}