Poj 2594 treasure transition ation (Transfer closure + minimum path overwrite)

// The build of the passed closure (Floyd) + the minimum path overwrite <br/> // This question is different from 1422 because it is a directed graph rather than Dag, the robot can go back to another path. <br/> // the roads of two different robots may contain some same point. <br/> // For example, 1-> 2, 2-> 3, 3-> 4, 4-> 2, 2-> 5. This graph can be completed by only one robot, however, if the passing closure is not set up <br/> // directly matches based on the edge, the result is incorrect, because the path overwrites that are usually referred, the vertex can only go through once at most, and this question can go through multiple times <br/> // Therefore, you must use Floyd to re-determine the connectivity and create a passing closure, create a New Graph Based on the closure, make sure that each vertex passes through only once <br/> // so that the minimum path overwrite can be obtained using the maximum matching method of the Bipartite Graph. <br/> # include <iostream> <B R/> using namespace STD; <br/> const int max = 502; <br/> int n, m; <br/> bool a [Max] [Max]; <br/> int X, Y, vis [Max], G [Max] [Max], ymatch [Max]; <br/> void Floyd () // evaluate the passed closure and create a new graph <br/>{< br/> for (int K = 1; k <= N; ++ K) <br/> for (INT I = 1; I <= N; ++ I) <br/> for (Int J = 1; j <= N; ++ J) <br/>{< br/> if (I = J) continue; <br/> if (a [I] [k] & A [k] [J]) <br/> A [I] [J] = 1; <br/>}< br/> for (INT I = 1; I <= N; ++ I) <br/> for (Int J = 1; J <= N; ++ J) <br/> if (a [I] [J]) g [I] [J] = 1; <br/>}< br/> bool findpath (int x) // search for augmented path <br/>{< br/> for (INT y = 1; Y <= y; ++ y) <br/>{< br/> If (! Vis [y] & G [x] [Y]) <br/>{< br/> vis [y] = 1; <br/> If (ymatch [y] =-1 | findpath (ymatch [y]) // If V is not matched or V finds a new augmented path <br/>{< br/> ymatch [y] = x; <br/> return true; <br/>}< br/> return false; <br/>}< br/> int maxmatch () // returns the maximum number of matches <br/>{< br/> int ans = 0; <br/> memset (ymatch,-1, sizeof (ymatch )); // initialization <br/> for (INT x = 1; x <= x; ++ X) // note the numbers <br/> {<br/> memset (VIS, 0, sizeof (VIS); <br/> If (findpath (x )) + + ans; <br/>}< br/> return ans; <br/>}< br/> int main () <br/>{< br/> int X, Y; <br/> while (scanf ("% d", & N, & M) & N) <br/>{< br/> memset (A, 0, sizeof (a); <br/> memset (G, 0, sizeof (g )); <br/> while (M --) <br/>{< br/> scanf ("% d", & X, & Y ); <br/> A [x] [Y] = 1; <br/>}< br/> Floyd (); <br/> X = y = N; <br/> printf ("% d/N", X-maxmatch (); <br/>}< br/> return 0; <br/>}