Clever graph creation with a minimum cut of zoj 3475

For the first time, contact this type of graph creation question and record it.

Question:In a 20*20 square, some walls are built so that X and some A and E are not connected to the outside world, and the construction of the wall has a corresponding cost. For a qualified, corresponding income.

First, 2 ^ 5 refers to which a is to be enclosed, and then the minimum cost is required. This is a typical maximum flow least cut problem. X and a listed are connected to the same edge with the INF traffic, and E and the same edge with the INF traffic.

The above is a concise question of watashi. If I haven't responded to the problem at the beginning

Why is the minimum cut after creating a graph the minimum cost of separating X from some A and E? NOTE: If an edge is selected as a cut edge, the wall between two grids in the square is selected.

Note that the S set is X and A, and the T set is E and boundary. Therefore, the vertex in the s set after the minimum cut cannot reach the t set,

Therefore, it completes the function to enclose X and a and separate them from E.

 

View code

# Include <cstdio> # Include <Cstring> # Include < Set > # Include <Algorithm> Using   Namespace  STD;  Const   Int Max = 405  ;  Const   Int INF = 1000000000  ;  Struct  { Int  V, C, next;} edge [  2000 ], Edge [ 2000  ];  Int  E, head [Max], S, T;  Int  Gap [Max], cur [Max];  Int  Pre [Max], DIS [Max];  Void Add_edge ( Int S, Int T, Int C) {edge [e]. v = T; edge [e]. c = C; edge [e]. Next = Head [s]; head [s] = E ++ ; Edge [e]. v = S; edge [e]. c = C; edge [e]. Next = Head [T]; head [T] = E ++ ;}  Int Min ( Int A, Int B ){ Return (A =- 1 | B <)?B: ;}  Int SAP ( Int S, Int T, Int  N) {memset (gap,  0 , Sizeof  (GAP); memset (DIS,  0 , Sizeof  (DIS ));  Int  I;  For (I =0 ; I <n; I ++) cur [I] = Head [I];  Int U = pre [s] = s, maxflow = 0 , Aug =- 1  , V; Gap [  0 ] = N;  While (DIS [s] < N) {loop:  For (I = cur [u]; I! =- 1 ; I = Edge [I]. Next) {v =Edge [I]. V;  If (Edge [I]. c> 0 & Dis [u] = dis [v] + 1  ) {Aug = Min (Aug, edge [I]. c); Pre [v] = U; cur [u] = I; u = V;  If (U = T ){  For (U = pre [u]; V! = S; V = u, u =Pre [u]) {edge [cur [u]. c -= Aug; edge [cur [u] ^ 1 ]. C ++ = Aug;} maxflow + = Aug; Aug =- 1  ;}  Goto  Loop ;}}  Int Mindis = N;  For (I = head [u]; I! =- 1 ; I = Edge [I]. Next) {v = Edge [I]. V;  If (Edge [I]. c> 0 & Dis [v] < Mindis) {cur [u] = I; mindis = Dis [v] ;}}  If (-- Gap [dis [u]) = 0 ) Break  ; Gap [dis [u] = Mindis + 1 ] ++ ; U = Pre [u];}  Return  Maxflow ;}  Int  X [Max], Y [Max], Z [Max], H [Max], n, m;  Int ID ( Int X, Int Y ){ Return X * m + Y ;}  Int  Main (){ While (Scanf ( "  % D  " , & N, & M )! = EOF) {s = N * m; t = n * m + 1  ; Memset (Head, - 1 , Sizeof  (Head); e = 0  ;  For ( Int I =0 , F; I <= N; I ++ ){  For ( Int J = 0 ; J <m; j ++ ) {Scanf (  "  % D  " ,& F); add_edge (I = 0 ? T: ID (I- 1 , J), I = n? T: ID (I, j), f );} If (I = N) Break  ;  For ( Int J = 0 , F; j <= m; j ++ ) {Scanf (  "  % D  " ,& F); add_edge (J = 0 ? T: ID (I, j- 1 ), J = m?T: ID (I, j), f );}}  Int  K; scanf (  "  % D  " , & K ); Int T = K;  For ( Int I = 0 ; I <t; I ++ ) {Scanf (  "  % D  " , & Z [I], & X [I], & Y [I]);  If (Z [I] = 0  ) {Swap (Z [I], Z [  0  ]); Swap (Y [I], Y [  0  ]); Swap (X [I], X [  0  ]);}  Else   If (Z [I] < 0 ) {Add_edge (ID (X [I], Y [I]), T, INF); I --; T -- ;}} Memcpy (edge, edge,  Sizeof  (Edge); memcpy (H, Head,  Sizeof  (Head ));  Int Ans = INF;  Int TMP = E;  For ( Int I = 1 , F =0 ; I <( 1 <T); I + = 2  ) {F = 0  ; Memcpy (Head, H,  Sizeof  (H); memcpy (edge, edge,  Sizeof (Edge); E = TMP;  For ( Int J = 0 ; J <t; j ++){  If (I &( 1 < J) {add_edge (S, ID (X [J], Y [J]), INF); f -= Z [J] ;}} F + = SAP (S, T, N * m + 2  ); Ans = Min (ANS, f);} printf (  "  % D \ n  "  , ANS );} Return   0  ;}