对于删除每个对(x,y), 可以发现他对答案的贡献为代数余子式$A_{xy}$
复习了一下线代后发现代数余子式可以通过伴随矩阵求出, 即$A_{xy}=A^*[y][x]$, 伴随矩阵$A^*=|A|A^{-1}$可以通过高斯消元$O(\frac{n^3}{\omega})$求出
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
//headconst int N = 2010;
bitset<2*N> g[N];
int x[N*N], y[N*N], n, m;int main() {scanf("%d%d", &n, &m);REP(i,1,n) g[i][i+n]=1;REP(i,1,m) {scanf("%d%d", x+i, y+i);g[x[i]][y[i]]=1;}REP(i,1,n) {REP(j,i,n) if (g[j][i]) {swap(g[i],g[j]);break;}REP(j,1,n) if (j!=i&&g[j][i]) g[j]^=g[i];}REP(i,1,m) puts(g[y[i]][x[i]+n]?"NO":"YES");
}
)
