Hit 2713 matrix1

Hit_2713

The biggest one is equal to the least one left behind. Therefore, the problem becomes that the precious stones with the minimum value are not adjacent to each other.

First, each grid is dyed in black and white Based on the parity of I + J, and then s is connected to all the black grids. The capacity is the value of the corresponding grid, and then all the white grids are connected to T, capacity also corresponds to the value of the grid, and finally connects the Black Grid and the adjacent white grid, the capacity is INF.

The expected solution is a cut in this figure. The points that can be reached from S represent the white grids left behind and the black grids taken away, the point that can reach t represents the black lattice left behind and the white lattice taken away. In order to minimize the value of the remaining lattice, you can perform a minimum cut on the source image.

# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Define Maxd 2510# Define Maxm 15010 # Define INF 0x3f3f3f Int  N, m, first [maxd], E, next [maxm], V [maxm], flow [maxm];  Int  Sum, S, T, d [maxd], Q [maxd], work [maxd];  Int DX [] = {- 1 , 1 , 0 , 0 }, Dy [] = { 0 , 0 ,-1 , 1  };  Void Add ( Int X, Int Y, Int  Z) {v [E] = Y, flow [e] = Z; next [E] = First [X], first [x] = e ++ ;}  Void  Init (){  Int  I, J, K, X, Ni, NJ; scanf ( "  % D  " , & N ,& M); s = 0 , T = n * m + 1  ; Memset (first, - 1 , Sizeof (First [ 0 ]) * (T + 1  ); E = 0  ; Sum =0  ;  For (I = 1 ; I <= N; I ++ )  For (J = 1 ; J <= m; j ++ ) {Scanf (  "  % D  " , & X), sum + = X;  If (I + J & 1 ) {Add (S, (I - 1 ) * M + J, x), add (I- 1 ) * M + J, S, 0  );  For (K = 0 ; K < 4 ; K ++ ) {Ni = I + dx [K], nj = J + Dy [k];  If (Ni> = 1 & Ni <= N & NJ> = 1 & NJ <= M) add (I - 1 ) * M + J, (Ni- 1 ) * M + NJ, INF), add (Ni- 1 ) * M + NJ, (I- 1 ) * M + J, 0  );}}  Else  Add (I - 1 ) * M + J, t, x), add (T, (I-1 ) * M + J, 0  );}}  Int  BFS (){  Int I, j, rear = 0  ; Memset (D, - 1 , Sizeof (D [ 0 ]) * (T + 1  ); D [s] = 0 , Q [rear ++] =S;  For (I = 0 ; I <rear; I ++ )  For (J = first [Q [I]; J! =- 1 ; J = Next [J])  If (Flow [J] & D [V [J] =- 1  ) {D [V [J] = D [Q [I] + 1 , Q [rear ++] = V [J];  If (V [J] = T) Return   1  ;}  Return   0  ;}  Int DFS ( Int Cur, Int  A ){  If (Cur = T)  Return  A;  For (Int & I = work [cur]; I! =- 1 ; I = Next [I])  If (Flow [I] & D [V [I] = d [cur] + 1  )  If ( Int T = DFS (V [I], STD: min (A, flow [I]) {flow [I] -= T, flow [I ^ 1 ] + = T;  Return T ;}  Return   0  ;}  Int  Dinic (){  Int Ans = 0  , T;  While  (BFS () {memcpy (work, first,  Sizeof (First [ 0 ]) * (T + 1  )); While (T = DFS (S, INF) ans + = T ;}  Return  Ans ;}  Void  Solve () {printf (  "  % D \ n  " , Sum- Dinic ());}  Int  Main (){  Int  T; scanf ( "  % D  " ,& T );  While (T -- ) {Init (); solve ();}  Return   0  ;}