Nyoj Matchmaker problem (stable marriage problem)

Source: Internet
Author: User

Baidu has a stable marriage problem. What else is the GS algorithm?

To why high-end things.

The result code is exactly the same as the previous question.

Drunk.

1#include <iostream>2#include <cstdio>3#include <cstdlib>4#include <cstring>5#include <string>6#include <queue>7#include <algorithm>8#include <map>9#include <iomanip>Ten#include <climits> One#include <string.h> A#include <cmath> -#include <stdlib.h> -#include <vector> the#include <Set> - #defineINF 1e7 - #defineMAXN 100010 - #defineMAXN 50 + #defineMAXM 1000 - #defineMod 1000007 + using namespacestd; AtypedefLong LongLL; at  -  - intT; - intu, v; - intN, M; - intans; invector<int> g[555]; - intvis[10010]; to intmatch[10010]; +  - BOOLPathintu) the { *      for(inti =0; I < g[u].size (); ++i) { $         intv =G[u][i];Panax Notoginseng         if(true= = Vis[v])Continue; -VIS[V] =true; the         if(Match[v] = =-1||path (Match[v])) { +MATCH[V] =u; A             return true; the         } +     } -     return false; $ } $  - voidHungarian () - { theAns =0; -memset (match,-1,sizeof(Match));Wuyi      for(inti =1; I <= N; ++i) { thememset (Vis,0,sizeof(Vis)); -         if(Path (i)) ans++; Wu     } - } About  $ voidRun () - { -scanf"%d%d",&n,&m); -      for(inti =0; I <= N; ++i) { A g[i].clear (); +     } the      for(inti =0; I < m; ++i) { -scanf"%d%d",&u,&v); $ G[u].push_back (v); the     } the Hungarian (); theprintf"%d\n", ans); the      - } in intMain () the { thescanf"%d", &T); About      while(t--) the run (); the     return 0; the}

Nyoj Matchmaker problem (stable marriage problem)

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