二分图求最大独立集模板

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 5000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;

const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

vector<int> G[N];
vector<int> rG[N];

int L[N], totL;
int R[N], totR;
bool vis[N];
int matchL[N], matchR[N];
bool visL[N], visR[N];
vector<int> V;

int path(int u) {
    visL[u] = true;
    for(auto &v : G[u]) {
        if(!visR[v]) {
            visR[v] = true;
            if(matchR[v] == 0 || path(matchR[v])) {
                matchR[v] = u;
                matchL[u] = v;
                return 1;
            }
        }
    }
    return 0;
}

int main() {
    
    int cnt = 0;
    for(int i = 1; i <= totL; i++) {
        memset(visL, 0, sizeof(visL));
        memset(visR, 0, sizeof(visR));
        if(path(i)) {
            cnt++;
        }
    }

    cnt = n - cnt;

    memset(visL, 0, sizeof(visL));
    memset(visR, 0, sizeof(visR));

    for(int i = 1; i <= totL; i++) {
        if(!matchL[i]) {
            path(i);
        }
    }

    for(int i = 1; i <= totL; i++) {
        if(visL[i]) {
            V.push_back(L[i]);
        }
    }

    for(int i = 1; i <= totR; i++) {
        if(!visR[i]) {
            V.push_back(R[i]);
        }
    }

    printf("%d\n", cnt);
    for(int i = 0; i < SZ(V); i++) {
        printf("%d%c", V[i], " \n"[i == SZ(V) - 1]);
    }

    return 0;
}

/*
*/