Poj 1087 a plug for Unix (network stream-maximum stream (maximum binary matching ))

// Network stream-use the largest stream in the diagram for big match <br/> // connect the device to the source point <br/> // socket (plug) connect to the sink <br/> // The adapter is connected to the socket, and the weight is INF <br/> // converts the largest matching problem diagram to the largest flow problem, get the answer by subtracting the largest stream from the aggregate stream of the sink point <br/> # include <iostream> <br/> # include <queue> <br/> # include <map> <br/> # include <cstring> <br/> # include <string> <br/> # define INF 2000000000 <br/> using namespace STD; <br/> int flow [505] [505], Cap [505] [505]; <br/> int rflow [505], pre [505]; <br/> int U, V, C, maxflow, St, Ed, n, m, K; <Br/> int node; <br/> int main () <br/> {<br/> // freopen ("in.txt", "r", stdin ); <br/> maxflow = 0; <br/> node = 1; // The node number, the electric appliance and plug are numbered accordingly. <br/> ST = 0; <br/> ED = 1; <br/> string plug, device, adapter; <br/> Map <string, int> m; // map is used to associate node numbers of various electrical appliances and adapters. <br/> CIN> N; <br/> memset (flow, 0, sizeof (flow); <br/> memset (Cap, 0, sizeof (CAP )); <br/> for (INT I = 0; I <n; ++ I) <br/>{< br/> CIN> plug; <br/> M [plug] = ++ n Ode; <br/> CAP [node] [ed] = 1; <br/>}< br/> CIN> m; <br/> for (INT I = 0; I <m; ++ I) <br/>{< br/> CIN> device> plug; <br/> M [device] = ++ node; <br/> CAP [st] [M [device] = 1; <br/> If (! M. count (plug) <br/>{< br/> M [plug] = ++ node; <br/>}< br/> CAP [M [device] [M [plug] = 1; <br/>}< br/> CIN> K; <br/> for (INT I = 0; I <K; ++ I) <br/>{< br/> CIN> adapter> plug; <br/> If (! M. Count (adapter) <br/>{< br/> M [adapter] =++ node; <br/>}< br/> If (! M. count (plug) <br/>{< br/> M [plug] = ++ node; <br/>}< br/> CAP [M [adapter] [M [plug] = inf; // note that, he also said that the number of adapters is infinite, so the weight for constructing the adapter to the plug should be INF <br/>}< br/> (;;) // maximum stream template <br/>{< br/> queue <int> q; <br/> memset (rflow, 0, sizeof (rflow )); <br/> rflow [st] = inf; <br/> q. push (ST); <br/> while (! Q. empty () <br/>{< br/> U = Q. front (); <br/> q. pop (); <br/> for (V = 0; v <= node; ++ v) <br/>{< br/> If (! Rflow [v] & Cap [u] [v]> flow [u] [v]) <br/> {<br/> rflow [v] = min (rflow [u], Cap [u] [v]-flow [u] [v]); <br/> pre [v] = u; <br/> q. push (V); <br/>}< br/> If (rflow [ed] = 0) break; <br/> for (u = Ed; u! = ST; u = pre [u]) <br/>{< br/> flow [pre [u] [u] + = rflow [ed]; // update forward traffic <br/> flow [u] [pre [u]-= rflow [ed]; // update reverse traffic <br/>}< br/> maxflow + = rflow [ed]; <br/>}< br/> printf ("% d/N ", m-maxflow); // The sum of the traffic to the sink point minus the maximum stream is the answer <br/> return 0; <br/>}