# Spoj 1693 coconuts

Source: Internet
Author: User

Spoj_1693 coconuts

There are two types of choices for each person. Based on their final choice, they can be divided into two sets. Obviously, if there is a friend between the sets, the edge should be deleted, this actually forms a cut, corresponding to the number of friends with different opinions. In order to reflect the cost of changing your opinion, you can connect S and all the people who start to select 1 to an edge with a capacity of 1, then, a person and T with a capacity of 1 will be connected, and the edge of the friend relationship will be a undirected edge with a capacity of 1.In this way, the final result can be obtained by finding the minimum cut for this graph.

`# Include <stdio. h> # Include < String . H> # Include <Algorithm> # Define Maxd 310 # Define Maxm 90610 # Define INF 0x3f3f3fInt  N, m, first [maxd], E, next [maxm], V [maxm], flow [maxm];  Int  S, T, d [maxd], Q [maxd], work [maxd];  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, X, Y; s = 0 , T = n + 1  ; Memset (first, - 1 , Sizeof (First [ 0 ]) * (T + 1  ); E = 0  ;  For (I = 1 ; I <= N; I ++) {Scanf (  "  % D  " ,& X );  If  (X) add (S, I,  1 ), Add (I, S, 0  );  Else  Add (I, T,  1 ), Add (t, I, 0  );} For (I = 0 ; I <m; I ++ ) {Scanf (  "  % D  " , & X ,& Y); add (x, y,  1 ), Add (Y, X, 1  );}}  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 ]) * (N + 1  ));  While (T = DFS (S, INF) ans + = T ;}  Return  Ans ;}  Void Solve () {printf (  "  % D \ n  "  , Dinic ());}  Int  Main (){  While (Scanf ( "  % D  " , & N, & M), N | M) {Init (); solve ();}  Return   0  ;} `

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