1182. Team Them Up!

Time limit: 1.0 second Memory limit: 64 MB

Your task is to divide a number of persons into two teams, in such a way, that:

• everyone belongs to one of the teams;
• every team has at least one member;
• every person in the team knows every other person in his team;
• teams are as close in their sizes as possible.

This task may have many solutions. You are to find and output any solution, or to report that the solution does not exist.

Input

For simplicity, all persons are assigned a unique integer identifier from 1 to N.

The first line contains a single integer number N (2 ≤ N ≤ 100) - the total number of persons to divide into teams, followed by N lines - one line per person in ascending order of their identifiers. Each line contains the list of distinct numbers A

_ij

 (1 ≤ A

_ij

 ≤ N, A

_ij

 ≠ i) separated by spaces. The list represents identifiers of persons that i

^th

 person knows. The list is terminated by 0.

Output

If the solution to the problem does not exist, then write a single message “No solution” (without quotes). Otherwise write a solution on two lines. On the first line write the number of persons in the first team, followed by the identifiers of persons in the first team, placing one space before each identifier. On the second line describe the second team in the same way. You may write teams and identifiers of persons in a team in any order.

Sample

input	output
52 3 5 01 4 5 3 01 2 5 01 2 3 04 3 2 1 0	3 1 3 52 2 4

Problem Author: Vladimir Kotov, Roman Elizarov

 

Problem Source: 2001-2002 ACM Northeastern European Regional Programming Contest

*****************************************************************************************************

求补图&&dfs算法&&背包问题f[][],,,染色问题，程序中有解释

*****************************************************************************************************

1 /*本题主要是要求补图  2 和dfs求连通块，每个连通块分成两部分x[][],y[][];  3 然后再用01背包求出最小值并标记  4 最后输出计算结果，此处注意x[i][0],y[i][0]是表示i个连通块中两部分的数量  5 v[i]数组用于标记，点i属于哪个组。  6 */  7 #include
        
           8 #include
         
            9 #include
          
            10 #include
           
             11 #include
            
              12 #include
             
               13 using namespace std; 14 const int maxn=110; 15 int e[1001][1001]; 16 int v[1001],x[1001][1001],y[1001][1001]; 17 int f[1001][1001]; 18 int n,m,j,i,p; 19 int tmp; 20 int absw(int x) 21 { 22 return (x<0)?(-x):x; 23 } 24 void init() 25 { 26 cin>>n; 27 memset(e,0,sizeof(e)); 28 for(i=0;i
              
               >tmp&&tmp) 31 { 32 e[i][tmp-1]=1; 33 } 34 } 35 for(i=0;i
               
                absw(j-maxn)) 93 ans=j;//最小差值 94 g1=0; 95 for(i=p-1;i>=0;i--)//背包问题从后往前找 96 if(f[i][ans]==0) 97 { 98 g1+=x[i][0]; 99 for(j=1;j<=x[i][0];j++)v[x[i][j]]=1;100 for(j=1;j<=y[i][0];j++)v[y[i][j]]=2;101 ans-=(x[i][0]-y[i][0]);102 }103 else104 {105 g1+=y[i][0];106 for(j=1;j<=x[i][0];j++)107 v[x[i][j]]=2;108 for(j=1;j<=y[i][0];j++)109 v[y[i][j]]=1;110 ans-=(y[i][0]-x[i][0]);111 }112 //输出计算结果113 cout<
                
                 <<' ';114 for(i=0;i
                
               
              
             
            
           
          
         
        

坚持！！！！！