hdu--5093--经典二分图的行列匹配

  1 #include <iostream>
  2 #include <cstring>
  3 using namespace std;
  4 
  5 int cntX , cntY , n , m;
  6 const int size = 55;
  7 char mp[size][size];
  8 int x[size][size];
  9 int y[size][size];
 10 bool vis[size*size];
 11 int linker[size*size];
 12 bool edge[size*size][size*size];
 13 
 14 void init( )
 15 {
 16     cntX = cntY = 1;
 17     memset( x , 0 , sizeof(x) );
 18     memset( y , 0 , sizeof(y) );
 19     memset( edge , false , sizeof(edge) );
 20     memset( linker , 0 , sizeof(linker) );
 21 }
 22 
 23 void solveX( )
 24 {
 25     for( int i = 1 ; i<=n ; i++ )
 26     {
 27         int j = 1;
 28         while( j<=m )
 29         {
 30             if( mp[i][j]==‘*‘ )
 31             {
 32                 x[i][j] = cntX;
 33                 ++ j;
 34                 while( j<=m )
 35                 {
 36                     if( mp[i][j]==‘*‘ )
 37                     {
 38                         x[i][j] = cntX;
 39                         ++ j;
 40                     }
 41                     else if( mp[i][j]==‘o‘ )
 42                     {
 43                         ++ j;
 44                         continue;
 45                     }
 46                     else
 47                     {
 48                         break;
 49                     }
 50                 }
 51                 ++ cntX;
 52             }
 53             else
 54             {
 55                 ++ j;
 56             }
 57         }
 58     }
 59 }
 60 
 61 void solveY( )
 62 {
 63     for( int i = 1 ; i<=m ; i++ )//列
 64     {
 65         int j = 1;
 66         while( j<=n )//行
 67         {
 68             if( mp[j][i]==‘*‘ )
 69             {
 70                 y[j][i] = cntY;
 71                 edge[ x[j][i] ][ y[j][i] ] = true;
 72                 ++ j;
 73                 while( j<=n )
 74                 {
 75                     if( mp[j][i]==‘*‘ )
 76                     {
 77                         y[j][i] = cntY;
 78                         edge[ x[j][i] ][ y[j][i] ] = true;
 79                         ++ j;
 80                     }
 81                     else if( mp[j][i]==‘o‘ )
 82                     {
 83                         ++ j;
 84                         continue;
 85                     }
 86                     else
 87                     {
 88                         break;
 89                     }
 90                 }
 91                 ++ cntY;
 92             }
 93             else
 94             {
 95                 ++ j;
 96             }
 97         }
 98     }
 99 }
100 
101 bool dfs( int u )
102 {
103     for( int v = 1 ; v<cntY ; v++ )
104     {
105         if( edge[u][v] && !vis[v] )
106         {
107             vis[v] = true;
108             if( !linker[v] || dfs( linker[v] ) )
109             {
110                 linker[v] = u;
111                 return true;
112             }
113         }
114     }
115     return false;
116 }
117 
118 int solve( )
119 {
120     int ans = 0;
121     for( int i = 1 ; i<cntX ; i++ )
122     {
123         memset( vis , false , sizeof(vis) );
124         if( dfs(i) )
125             ++ ans;
126     }    
127     return ans;
128 }
129 
130 int main()
131 {
132     cin.sync_with_stdio(false);
133     int t , ans;
134     while( cin >> t )
135     {
136         while( t-- )
137         {
138             init( );
139             cin >> n >> m;
140             for( int i = 1 ; i<=n ; i++ )
141             {
142                 cin >> (mp[i]+1);
143             }
144             solveX( );
145             solveY( );
146             /*
147             for( int i = 1 ; i<=n ; i++ )
148             {
149                 for( int j = 1 ; j<=m ; j++ )
150                 {
151                     cout << x[i][j] << " ";
152                 }
153                 cout << endl;
154             }
155             for( int i = 1 ; i<=m ; i++ )
156             {
157                 for( int j = 1 ; j<=n ; j++ )
158                 {
159                     cout << y[i][j] << " ";
160                 }
161                 cout << endl;
162             }
163             */
164             ans = solve( );
165             cout << ans << endl;
166         }
167     }
168     return 0;
169 }