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=110;
const int INF=0x3f3f3f3f;
const int MOD=9973,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-6;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=0;i<G[u].size();i++){
        int v=G[u][i];
        if(color[v]==color[u])return false;
        if(!color[v]){
            color[v]=3-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,0);
        for(i=1;i<=n;i++)G[i].clear();
        for(i=1;i<=n;i++){
            while(scanf("%d",&a) && a)g[i][a]=1;
        }
        for(i=1;i<=n;i++){
            for(j=i+1;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,0);
        int ok=1;
        for(i=1;i<=n;i++){
            if(!color[i]){
                color[i]=1;
                if(!bipartite(i)){
                    ok=0;
                    break;
                }
            }
        }

        printf("%s
",ok?"YES":"NO");
    }
    return 0;
}