poj 1419 Graph Coloring 【最大团】

最新推荐文章于 2021-11-06 17:14:22 发布

原创最新推荐文章于 2021-11-06 17:14:22 发布 · 805 阅读

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最大匹配(权值匹配)、独立集、团 && 最小路径、点覆盖专栏收录该内容

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本文探讨了在给定节点和边的无向图中，如何通过应用黑白两种颜色进行最优着色，使得黑色节点数量最大化，同时确保相邻节点颜色不同。通过输入描述的实例，演示了解决该问题的方法，并提供了AC代码实现。

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Graph Coloring

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4559		Accepted: 2085		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


题意：给定N个点M条无向边，现在对节点染黑、白色，要求相邻节点颜色不同。问最多可以把多少节点染成黑色,并输出任意一组方案。

思路：求解二分图的最大独立集 = 补图的最大团。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXN (100+10)
using namespace std;
int N, M;
int Clique[MAXN];//Clique[i]记录(i-N)这些节点可以构成的最大团
int Map[MAXN][MAXN];
int New[MAXN][MAXN];//每次建立从i-N这些节点  DFS
int used[MAXN], Rec[MAXN];
int ans;
int DFS(int T, int cnt)//DFS遍历层次 计数
{
    if(T == 0)
    {
        if(ans < cnt)
        {
            ans = cnt;
            memcpy(Rec, used, sizeof(used));
            return 1;
        }
        return 0;
    }
    for(int i = 0; i < T; i++)
    {
        if(T - i + cnt <= ans) return 0;
        int u = New[cnt][i];
        if(Clique[u] + cnt <= ans) return 0;
        int num = 0;
        for(int j = i+1; j < T; j++)
            if(Map[u][New[cnt][j]] == 0)
                New[cnt+1][num++] = New[cnt][j];
        used[cnt+1] = u;
        if(DFS(num, cnt+1)) return 1;
    }
    return 0;
}
void MaxClique()
{
    memset(Clique, 0, sizeof(Clique));
    ans = 0;
    for(int i = N; i >= 1; i--)
    {
        used[1] = i; int Size = 0;
        for(int j = i+1; j <= N; j++)//根据后面的节点构建新图
            if(Map[i][j] == 0)
                New[1][Size++] = j;
        DFS(Size, 1);
        Clique[i] = ans;
    }
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &N, &M);
        memset(Map, 0, sizeof(Map));
        for(int i = 1; i <= M; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            Map[a][b] = Map[b][a] = 1;
        }
        MaxClique();
        printf("%d\n", ans);
        for(int i = 1; i <= ans; i++)
        {
            if(i > 1)
                printf(" ");
            printf("%d", Rec[i]);
        }
        printf("\n");
    }
    return 0;
}