Graph Coloring（ DP优化的求最大团模板题：求原图的最大独立集和输出集合元素可转化为求补图的最大团顶点数+输出最大团元素）

link：http://poj.org/problem?id=1419

Graph Coloring Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4508   Accepted: 2063   Special Judge Description You are to write a program that tries to find an optimal coloring for a given graph. Colors are applied to the nodes of the graph and the only available colors are black and white. The coloring of the graph is called optimal if a maximum of nodes is black. The coloring is restricted by the rule that no two connected nodes may be black.    Figure 1: An optimal graph with three black nodes  Input The graph is given as a set of nodes denoted by numbers 1...n, n <= 100, and a set of undirected edges denoted by pairs of node numbers (n1, n2), n1 != n2. The input file contains m graphs. The number m is given on the first line. The first line of each graph contains n and k, the number of nodes and the number of edges, respectively. The following k lines contain the edges given by a pair of node numbers, which are separated by a space. Output The output should consists of 2m lines, two lines for each graph found in the input file. The first line of should contain the maximum number of nodes that can be colored black in the graph. The second line should contain one possible optimal coloring. It is given by the list of black nodes, separated by a blank. Sample Input 1 6 8 1 2 1 3 2 4 2 5 3 4 3 6 4 6 5 6 Sample Output 3 1 4 5 Source Southwestern European Regional Contest 1995

AC code：

#include
#include
#define N 1010
/*
最大团 = 补图G的最大独立集数
———>最大独立集数 = 补图G'最大团
*/
//最大团模板
bool a[N][N];//a为图的邻接表(从1开始) 
int ans, cnt[N], group[N], n, m, vis[N];//ans表示最大团，cnt[N]表示当前最大团的节点数，group[N]用以寻找一个最大团集合 
bool dfs( int u, int pos )//u为当从前顶点开始深搜，pos为深搜深度（即当前深搜树所在第几层的位置） 
{
    int i, j;
    for( i = u+1; i <= n; i++)//按递增顺序枚举顶点 
	{
        if( cnt[i]+pos <= ans ) return 0;//剪枝 
        if( a[u][i] ) 
		{
             // 与目前团中元素比较，取 Non-N(i) 
            for( j = 0; j < pos; j++ ) if( !a[i][ vis[j] ] ) break; 
            if( j == pos )
			{     // 若为空，则皆与 i 相邻，则此时将i加入到 最大团中 
                vis[pos] = i;//深搜层次也就是最大团的顶点数目，vis[pos] = i表示当前第pos小的最大团元素为i（因为是按增顺序枚举顶点 ） 
                if( dfs( i, pos+1 ) ) return 1;    
            }    
        }
    }    
    if( pos > ans )
	{
            for( i = 0; i < pos; i++ )
                group[i] = vis[i]; // 更新最大团元素 
            ans = pos;
            return 1;    
    }    
    return 0;
} 
void maxclique()//求最大团 
{
    ans=-1;
    for(int i=n;i>0;i--)
    {
        vis[0]=i;
        dfs(i,1);
        cnt[i]=ans;
    }
}
int main()
{
	//freopen("D:\in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while( T-- )
	{
        scanf("%d%d",&n,&m );
        int x, y;
        memset( a, 0, sizeof(a));
        for(int i = 0; i < m; i++)
		{
            scanf("%d%d",&x,&y);
            a[x][y] = a[y][x] = 1;
        }
        //相邻顶点间有边相连，模型转换成求 无向图 最大独立集。 
		//要求原图的最大独立集，转化为求原图的补图的最大团(最大团顶点数量 = 补图的最大独立集)    
        for(int i = 1; i <= n; i++)//求原图的补图 
            for(int j = 1; j <= n; j++)
                if( i == j ) a[i][j] = 0;
                else    a[i][j] ^= 1;
        maxclique();//求最大团 
        if( ans < 0 ) ans = 0;//ans表示最大团
        printf("%d\n", ans );
        for(int i = 0; i < ans; i++)
            printf( i == 0 ? "%d" : " %d", group[i]);//group[N]用以寻找一个最大团集合 
        if( ans > 0 ) puts("");
    }        
}