Poj 1087 a plug for Unix (max stream, EK algorithm)

Question: Reading is disgusting ···. That is to say, there are now n Different jacks, m different appliances, and K adapters. (The adapter is equivalent to an intermediate socket, for example, an adapter is (x, y ). There is an electrical appliance that must be inserted with an X Jack, but now there is only one y Jack, then it can be connected through an adapter ). In addition, you must note that the number of adapters is provided, and the number of each adapter is unlimited.
Question: the general idea of creating a graph. S --> Electrical Appliance --> adapter --> Jack --> T

# Include <queue> # include <cstring> # include <iostream> using namespace STD; # define n 101 # define INF 9999 char Recep [N] [25]; char device [N] [25]; struct item {char in [25]; Char out [25];} adapt [N]; bool vis [N * 3]; int pre [N * 3]; int CAP [N * 3] [N * 3]; int flow [N * 3] [N * 3]; int n, m, k; int S, T, Res; void build_map () {S = 0; t = n + M + k + 1; memset (Cap, 0, sizeof (CAP )); int I, j; for (I = 1; I <= m; I ++) {CAP [s] [I] = 1; // s to power For (j = 1; j <= N; j ++) // electrical appliance to Jack if (! Strcmp (device [I], Recep [J]) CAP [I] [J + M + k] = 1; for (j = 1; j <= K; j ++) // electrical appliance to adapter if (! Strcmp (device [I], adapt [J]. in) CAP [I] [J + M] = 1 ;}for (I = 1; I <= K; I ++) {for (j = 1; j <= K; j ++) // if (I! = J &&! Strcmp (adapt [I]. out, adapt [J]. in) CAP [I + M] [J + M] = inf; // The number is not limited, so it is inffor (j = 1; j <= N; j ++) // adapter to Jack if (! Strcmp (adapt [I]. out, Recep [J]) CAP [I + M] [J + M + k] = 1 ;}for (I = 1; I <= N; I ++) // Jack to TCAP [I + M + k] [T] = 1;} bool find_path () // BFS find augmented path {memset (PRE,-1, sizeof (-1); memset (VIS, 0, sizeof (VIS); queue <int> que; vis [s] = 1; que. push (s); While (! Que. empty () {int u = que. front (); que. pop (); For (INT v = T; V> = s; V --) {If (! Vis [v] & Cap [u] [v]> flow [u] [v]) {vis [v] = 1; Pre [v] = u; if (V = T) return true; que. push (v) ;}}return false;} int max_flow () {res = 0; memset (flow, 0, sizeof (flow); While (1) {If (! Find_path () break; int TT = T; int minflow = inf; while (pre [TT]! =-1) {If (minflow> CAP [pre [TT] [TT]-flow [pre [TT] [TT]) minflow = CAP [pre [TT] [TT]-flow [pre [TT] [TT]; TT = pre [TT];} TT = T; while (pre [TT]! =-1) {flow [pre [TT] [TT] + = minflow; flow [TT] [pre [TT]-= minflow; tt = pre [TT];} res + = minflow;} return res;} int main () {int I; char name [25]; scanf ("% d ", & N); for (I = 1; I <= N; I ++) scanf ("% s", Recep [I]); scanf ("% d ", & M); for (I = 1; I <= m; I ++) scanf ("% S % s", name, device [I]); scanf ("% d", & K); for (I = 1; I <= K; I ++) scanf ("% S % s", adapt [I]. in, adapt [I]. out); build_map (); printf ("% d \ n", M-max_flow (); Return 0 ;}