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HDU 4751 Divide Groups（二分圖的判斷）

編輯：C++入門知識

HDU 4751 Divide Groups（二分圖的判斷）

Problem Description
This year is the 60th anniversary of NJUST, and to make the celebration more colorful, Tom200 is going to invite distinguished alumnus back to visit and take photos.
After carefully planning, Tom200 announced his activity plan, one that contains two characters:
1. Whether the effect of the event are good or bad has nothing to do with the number of people join in.
2. The more people joining in one activity know each other, the more interesting the activity will be. Therefore, the best state is that, everyone knows each other.
The event appeals to a great number of alumnus, and Tom200 finds that they may not know each other or may just unilaterally recognize others. To improve the activities effects, Tom200 has to divide all those who signed up into groups to take part in the activity at different time. As we know, one's energy is limited, and Tom200 can hold activity twice. Tom200 already knows the relationship of each two person, but he cannot divide them because the number is too large.
Now Tom200 turns to you for help. Given the information, can you tell if it is possible to complete the dividing mission to make the two activity in best state.
Input The input contains several test cases, terminated by EOF.
Each case starts with a positive integer n (2<=n<=100), which means the number of people joining in the event.
N lines follow. The i-th line contains some integers which are the id
of students that the i-th student knows, terminated by 0. And the id starts from 1.
Output If divided successfully, please output "YES" in a line, else output "NO".

Sample Input

Sample Output

YES

題意：n個人分成兩個集合，要每一個圖的認識關系都是完全圖，是否可以分成

思路：先把不是相互認識的放在一個連接表聯系起來，這些人必須不是一個集合的，然後染色法判斷

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i < b; i++)
#define free(i,b,a) for(i = b; i >= a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 105

vectorg[N];
int n,vis[N];
int a[N][N];

bool dfs(int x,int color)
{
     int i;
     vis[x]=color;
     fre(i,0,g[x].size())
	{
        int to=g[x][i];

        if(vis[to]!=-1)
		{
			if(vis[to]==color) return false;  //不認識但是在一個集合
			continue;
		}

        if(!dfs(to,!color))  //不認識就得在不同的集合
		  return false;
	}
   return true;
}

bool solve()
{
	int i,j;
	mem(vis,-1);

	fre(i,1,n+1)
	{
		if(vis[i]==-1&&dfs(i,0)==false)
			return false;
	}
   return true;

}

int main()
{
	int i,j;
	while(~sf(n))
	{
		fre(i,1,n+1)
		  g[i].clear();

		int x;
		mem(a,0);
		fre(i,1,n+1)
		{
			while(sf(x),x)
				{
				  a[i][x]=1;
				}
		}

		fre(i,1,n+1)
		  fre(j,i+1,n+1)
		   if(a[i][j]==0||a[j][i]==0)
		   {
		   	  g[i].push_back(j);
		   	  g[j].push_back(i);
		   }

	   if(solve())
		pf("YES\n");
	   else
		pf("NO\n");
	}
     return 0;
}