二分图匹配学习——匈牙利算法模板

DFS(邻接矩阵)

const int MAXN=1000;
int p,n;  //u,v数目
int g[MAXN][MAXN];//左右集合连接情况
int linker[MAXN];
bool used[MAXN];
bool dfs(int u)
{
    int v;
    for(v=1; v<=n; v++)
        if(g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=1; u<=p; u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))
            res++;
    }
    return res;
}

DFS(邻接表)

const int MAXN=10050;
int linker[MAXN];
bool used[MAXN];
vectorg[MAXN];

int n;

bool dfs(int u)
{
    for(int i=0; i

BFS

const int MAXN = 1000;
int g[MAXN][MAXN];
int Mx[MAXN], My[MAXN], p, n;
int chk[MAXN];
int Q[MAXN];
int pre[MAXN];
int hungary()
{
    int res = 0;
    int qs, qe;
    memset(Mx, -1, sizeof(Mx));
    memset(My, -1, sizeof(My));
    memset(chk, -1, sizeof(chk));
    for (int i = 1; i <= p; i++)
    {
        if (Mx[i] == -1)  //对于x集合中的每个没有匹配的点i进行一次bfs找交错轨
        {
            qs = qe = 0;
            Q[qe++] = i;
            pre[i] = -1;
            bool flag = 0;//判断是否找到
            while (qs < qe && !flag)
            {
                int u = Q[qs];
                for (int v = 1; v <= n && !flag; v++)
                    if (g[u][v]//如果u和v相连
                            && chk[v] != i)//并且v没有被u check过
                    {
                        chk[v] = i;
                        Q[qe++] = My[v];//放进
                        if (My[v] >= 0)//如果v和其他的相连,则修改之
                            pre[My[v]] = u;
                        else  //直到找到一个u和v都没有用过的
                        {
                            flag = 1;
                            int d = u, e = v;
                            while (d != -1)  //确保回到最初
                            {
                                int t = Mx[d];
                                Mx[d] = e;
                                My[e] = d;
                                d = pre[d];
                                e = t;
                            }
                        }
                    }
                qs++;
            }
            if (Mx[i] != -1)
                res++;
        }
    }
    return res;
}

Hopcroft-Carp 算法

#define MAXN 128

const int INF = 1 << 28;

int g[MAXN][MAXN], Mx[MAXN], My[MAXN], Nx, Ny;
int dx[MAXN], dy[MAXN], dis;
bool vst[MAXN];

bool searchP(void)
{
	queue Q;
	dis = INF;
	memset(dx, -1, sizeof(dx));
	memset(dy, -1, sizeof(dy));
	for (int i = 0; i < Nx; i++)
	{
		if (Mx[i] == -1)
		{
			Q.push(i);
			dx[i] = 0;
		}
	}
	while (!Q.empty())
	{
		int u = Q.front(); Q.pop();
		if (dx[u] > dis)
			break;
		for (int v = 0; v < Ny; v++)
		{
			if (g[u][v] && dy[v] == -1)
			{
				dy[v] = dx[u]+1;
				if (My[v] == -1)
					dis = dy[v];
				else
				{
					dx[My[v]] = dy[v]+1;
					Q.push(My[v]);
				}
			}
		}
	}
	return dis != INF;
}

bool DFS(int u)
{
	for (int v = 0; v < Ny; v++)
	{
		if (!vst[v] && g[u][v] && dy[v] == dx[u]+1)
		{
			vst[v] = 1;
			if (My[v] != -1 && dy[v] == dis)
				continue;
			if (My[v] == -1 || DFS(My[v]))
			{
				My[v] = u;
				Mx[u] = v;
				return 1;
			}
		} 
	}
	return 0;
}

int MaxMatch(void)
{
	int res = 0;
	memset(Mx, -1, sizeof(Mx));
	memset(My, -1, sizeof(My));
	while (searchP())
	{
		memset(vst, 0, sizeof(vst));
		for (int i = 0; i < Nx; i++)
		{
			if (Mx[i] == -1 && DFS(i))//if (dx[i] == 0 && DFS(i))
				res++;
		}
	}
	return res;
}