HDU-4751 Divide Groups 染色问题

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4751

　　题意：有n个人，每个人都认识一些人，要求把他们分成两个集合，使得两个集合中的人都相符两两认识。

　　如果两个人单向认识或者相互不认识，那么必定在不同的集合，因此建立边，染色就可以了。。

 //STATUS:C++_AC_31MS_388KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 #pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e60;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int n;
 int g[N][N],color[N];
 vector<int> G[N];

 bool bipartite(int u)
 {
     for(int i=;i<G[u].size();i++){
         int v=G[u][i];
         if(color[v]==color[u])return false;
         if(!color[v]){
             color[v]=-color[u];
             if(!bipartite(v))return false;
         }
     }
     return true;
 }

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,a,b;
     while(~scanf("%d",&n))
     {
         mem(g,);
         for(i=;i<=n;i++)G[i].clear();
         for(i=;i<=n;i++){
             while(scanf("%d",&a) && a)g[i][a]=;
         }
         for(i=;i<=n;i++){
             for(j=i+;j<=n;j++){
                 if( (!g[i][j]&&g[j][i]) || (g[i][j]&&!g[j][i]) || (!g[i][j]&&!g[j][i])) {
                     G[i].push_back(j);
                     G[j].push_back(i);
                 }
             }
         }

         mem(color,);
         int ok=;
         for(i=;i<=n;i++){
             if(!color[i]){
                 color[i]=;
                 if(!bipartite(i)){
                     ok=;
                     break;
                 }
             }
         }

         printf("%s\n",ok?"YES":"NO");
     }
     return ;
 }