HDU 3829 Cat VS Dog

有p个小朋友参观动物园，动物园里面有两种动物，分别为猫和狗。规定一个小朋友喜欢猫就讨厌狗，喜欢狗就讨厌猫。

现在管理员要移走0一些动物，当然，移走也是有条件的。比如一个小朋友喜欢猫3，讨厌狗4.那么移走狗4，这个小朋友就会非常开心。同样，如果移走猫3，小朋友就会很不高兴。现在问怎么样才能使开心的小朋友的人数最多

二分图最大独立集

#include 
#include 
#include 
#include 
#define rep(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

int n, m, k, edge[509][509], done[509], link[509], f[1009][3];

bool dfs (int x)
{
    rep (i, 1, n)
        if (!done[i] && edge[x][i])
        {
            done[i] = 1;
            if (link[i] == -1 || dfs (link[i]))
            {
                link[i] = x;
                return 1;
            }
        }
    return 0;
}

int solve ()
{
    int ret = 0;
    rep (i, 1, n)
    {
        memset (done, 0, sizeof (done));
        if (dfs (i))
            ret++;
    }
    return ret;
}

int main ()
{
    while (scanf ("%d%d%d", &m, &k, &n) == 3)
    {
        memset (link, -1, sizeof (link));
        memset (edge, 0, sizeof (edge));
        memset (f, 0, sizeof (f));
        rep (i, 1, n)
        {
            rep (j, 1, 2)
            {
                getchar ();
                char now = getchar ();
                scanf ("%d", &f[i][j]);
                if (now == 'D')
                    f[i][j] += m;
            }
        }
        //rep (i, 1, n) printf ("%d === %d\n", f[i][1], f[i][2]);
        rep (i, 1, n)
            rep (j, 1, n)
                if (f[i][2] == f[j][1] || f[i][1] == f[j][2])
                    edge[i][j] = edge[j][i] = 1;
        //rep (i, 1, n) rep (j, 1, n) if (edge[i][j]) printf ("edge %d %d\n", i, j);
        cout << n - solve () / 2 << endl;
    }
    return 0;
}