Codeforces 277 A Learning Languages "DFS"

n individuals, each person will have some language, two people can communicate in the same language, and ask how many languages, at least, must be learned in order for the N-people to communicate.

First, according to the language of the N people they will, build the side

Dfs find out how many of the unicom block ans, plus ans-1 edge can let them connect

Pay attention to the case of everyone, 0 languages.

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5#include <vector>6 using namespacestd;7 8 Const intMAXN = the;9 intG[MAXN][MAXN],A[MAXN][MAXN];Ten intVIS[MAXN]; One intn,m; A  - voidDfsintu) { -Vis[u] =1; the      for(intv =1; v <= n;v++){ -         if(!vis[v] &&G[u][v]) Dfs (v); -     } - } +  - intMain () { +     intAns =0; Ascanf"%d%d",&n,&m); atMemset (G,0,sizeof(g)); -Memset (A,0,sizeof(a)); -     inttot =0; -      for(intU =1; u <= n;u++){ -         intK; -scanf"%d",&k); in         if(k = =0) tot++; -          for(inti =0; I < k;i++){ to             intv; +scanf"%d",&v); -A[U][V] =1; the         } *     } $     Panax Notoginseng     if(tot = n) printf ("%d\n", tot); -     Else { the              for(intU =1; u <= n;u++){ +              for(intv = u+1; v <= n;v++){ A              for(intK =1; k <= m;k++){ the                 if(A[u][k] && a[v][k]) g[u][v] = G[v][u] =1; +             } -             } $       } $      -      for(inti =1; I <= n;i++) { -         if(!Vis[i]) { theans++; - DFS (i);Wuyi         } the     } -printf"%d\n", ans-1); Wu     } -     return 0; About}

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Codeforces 277 A Learning Languages "DFS"