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本文介绍了一种基于二分图匹配和强连通分量(SCC)算法的实现，通过使用C++编程语言，详细展示了如何解决特定类型的二分图匹配问题。该算法利用深度优先搜索(DFS)来确定图中的强连通分量，并判断二分图是否能完美匹配。

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>

using namespace std;

const int M = 2005;
int ins[M];
int dfn[M];
int low[M];
int sccf[M];
vector<int>que[M];
stack<int>s;
int index, cont;
int flag;
int n, m;

void init() {

    int a, b, c, d;
    index = 1;
    cont = 0;
    for(int i = 0; i < n*2; i++) {

        que[i].clear();
    }
    for(int i = 0; i < m; i++) {

        scanf("%d%d%d%d", &a, &b, &c, &d);
        int u = a*2 + c;
        int v = b*2 + d;
        que[u].push_back(v^1);
        que[v].push_back(u^1);
    }
}

void Tanjian(int u) {

    int v;
    dfn[u] = low[u] = index++;
      s.push(u);
      ins[u] = 1;
    for(int i = 0; i < que[u].size(); i++) {

        v = que[u][i];
        if(!dfn[v]) {

            Tanjian(v);
            low[u] = min(low[u], low[v]);
        }
        else if(ins[v]) {

            low[u] = min(low[u], low[v]);
        }
    }
    if(dfn[u] == low[u]) {

        cont++;
        do {

            v = s.top();
            s.pop();
            sccf[v] = cont;
            ins[v] = 0;
        }while(v != u);
    }
}

void solve() {

    memset(ins, 0, sizeof(ins));
    memset(low, 0, sizeof(low));
    memset(dfn, 0, sizeof(dfn));
    memset(sccf, 0, sizeof(sccf));

    for(int i = 0; i < 2*n; i++) {

        if(!dfn[i]) {

            Tanjian(i);
        }
    }
    flag = 1;
    for(int i = 0; i < 2*n; i+= 2) {

        if(sccf[i] == sccf[i^1]) {

            flag = 0;
            break;
        }
    }
    if(flag)
        printf("YES\n");
    else
        printf("NO\n");

}

int main()
{
  while(scanf("%d", &n) != EOF) {

    scanf("%d", &m);
    init();
    solve();
  }
    return 0;
}