二分图最大匹配(匈牙利算法) POJ 3020 Antenna Placement

题目传送门

 /*
     题意：*的点占据后能顺带占据四个方向的一个*，问最少要占据多少个
     匈牙利算法：按坐标奇偶性把*分为两个集合，那么除了匹配的其中一方是顺带占据外，其他都要占据
 */
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 using namespace std;

 const int MAXN = 4e2 + ;
 const int INF = 0x3f3f3f3f;
 char s[][];
 int ha[][];
 bool vis[MAXN];
 int lk[MAXN];
 vector<int> G[MAXN];
 int dx[] = {-, , , };
 int dy[] = {, , -, };
 int un, vn;

 bool DFS(int u)
 {
     for (int i=; i<G[u].size (); ++i)
     {
         int v = G[u][i];
         if (!vis[v])
         {
             vis[v] = true;
             if (lk[v] == - || DFS (lk[v]))
             {
                 lk[v] = u;  return true;
             }
         }
     }

     return false;
 }

 int hungary(void)
 {
     int res = ;    memset (lk, -, sizeof (lk));
     for (int i=; i<=un; ++i)
     {
         memset (vis, false, sizeof (vis));
         if (DFS (i))    res++;
     }

     return res;
 }

 int main(void)        //POJ 3020 Antenna Placement
 {
     //freopen ("POJ_3020.in", "r", stdin);

     int t;  scanf ("%d", &t);
     while (t--)
     {
         int h, w;   scanf ("%d%d", &h, &w);
         for (int i=; i<=h; ++i)
         {
             scanf ("%s", s[i] + );
         }

         un = vn = ;
         for (int i=; i<=h; ++i)
         {
             for (int j=; j<=w; ++j)
             {
                 if (s[i][j] == '*')
                 {
                     if ((i+j) & )  ha[i][j] = ++un;
                     else    ha[i][j] = ++vn;
                 }
             }
         }

         for (int i=; i<=un; ++i)   G[i].clear ();

         for (int i=; i<=h; ++i)
         {
             for (int j=; j<=w; ++j)
             {
                 if (s[i][j] == '*' && (i+j) & )
                 {
                     for (int k=; k<; ++k)
                     {
                         int tx = i + dx[k]; int ty = j + dy[k];
                         if (tx >=  && tx <= h && ty >=  && ty <= w && s[tx][ty] == '*')
                             G[ha[i][j]].push_back (ha[tx][ty]);
                     }
                 }
             }
         }

         printf ("%d\n", un + vn - hungary ());
     }

     return ;
 }