pku Graph Coloring

最新推荐文章于 2025-07-01 08:10:11 发布

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最新推荐文章于 2025-07-01 08:10:11 发布

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分类专栏：最大团

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本文探讨了图论中的图染色问题，旨在通过一种算法寻找最优解，使得图中节点着色（仅限黑色和白色），且相邻节点颜色不同的情况下，尽可能多地使用黑色。文章提供了具体的输入输出示例及实现该目标的C++代码。

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Graph Coloring

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3352		Accepted: 1497		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

Sample Output

3
1 4 5

代码：

#include <cstdio>
#include <string>
#include<string.h>
#define NMAX 110
bool path[NMAX][NMAX];
int n, mmax;
int dp[NMAX];
bool v[NMAX];
int seq[NMAX], seq_pos;
//seq记录最大团集合
bool dfs(int pos, int size)
{
    int i, j, unvis;
    bool tv[NMAX];
    unvis = 0;
    for (i=pos; i<n; i++)
    {
        if (!v[i])
        {
            unvis ++;
        }
    }
    if (unvis == 0)  //|U| = 0
    {
        if (size > mmax)
        {
            mmax = size;
            seq_pos = 0;
            seq[ seq_pos ++] = pos+1;
            return true;
        }
        return false;
    }
    for (i=pos; i < n && unvis > 0 ; i++)
    {
        if (!v[i])
        {
            if (unvis + size <= mmax || dp[i] + size <= mmax)
            {
                return false;
            }
            v[i] = true;//U = U\{vi}
            unvis --;
            memcpy(tv, v, sizeof(v));
            for (j=0; j<n; j++) //U ∩N(vi);
            {
                if (!path[i][j])
                {
                    v[j] = true;
                }
            }
            if ( dfs(i, size+1) )
            {
                seq[ seq_pos ++] = pos+1;
                return true;
            }
            memcpy(v, tv, sizeof(v));
        }
    }//while U is not empty
    return false;
}
int max_clique()
{
    int i,j;
    mmax = 0;
    for (i=0; i<n; i++)
    {
        path[i][i] = false;
    }
    for (i=n-1; i>=0; i--)
    {
        for (j=0; j<n; j++) //Si ∩N(vi);
        {
            v[j] = !path[i][j];
        }
        dfs(i, 1);
        dp[i] = mmax;
    }
    return mmax;
}
int main()
{
    int i,j,x,y,e;
    int m,tn;
    scanf("%d", &m);
    while (m --)
    {
        scanf("%d %d", &n, &e);
        memset(path,0,sizeof(path));
        for (i=0; i<e; i++)
        {
            scanf("%d %d", &x,&y);
            x--;
            y--;
            path[x][y] = path[y][x] = true;
        }
//max independent set in original graph
//max clique in inverse graph
        for (i=0; i<n; i++)
        {
            for (j=0; j<n; j++)
            {
                path[i][j] = !path[i][j];
            }
        }
        memset(dp,0,sizeof(dp));
        printf("%d\n", max_clique());
        printf("%d", seq[0]);
        for (i=1; i<seq_pos; i++)
        {
            printf(" %d", seq[i]);
        }
        printf("\n");
    }
}