2019-02-24-Graph Coloring-POJ1419-图着色问题-最大独立集-最大团问题_最大割问题归约到图着色问题-优快云博客

本文详细解析了POJ-1419 Graph Coloring问题，介绍了如何利用最大团模板解决图中节点着色问题，确保不相连的节点可以使用相同的颜色，目标是最大量地使用黑色进行着色。

POJ-1419-Graph Coloring

Problem 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

题目简述

对于给定的图G(V,E)，用两种颜色（黑，白）进行着色，找出符合条件且黑色尽可能多的图，并给出其中一种方案。

条件：对于任意两个相邻的点，不能同时为黑色。

分析：题意即找出最大的集合，其中任意两点都不相邻（最大独立集，独立集），即找到补图的最大团。

条目检索：最大独立集，独立集，补图，最大团问题

输入输出

输入：无难点

输出：要求记录最大团集合（不要求排序）

使用模板-最大团(dfs)模板

//最大团模板
//dfs+dp+剪枝
struct MAX_CLIQUE{
    static const int N=110;
    bool G[N][N];    //G(V,E)记录数组
    int n,Max[N],Alt[N][N],ans;
    //Max[i]为集合{i,i+1,...,n}中最大团的数目，用于剪枝
    //Alt[i][j]，第i次dfs中当前可选集合的第j个元素
    //ans，答案记录
    bool DFS(int cur,int tot){
        if(!cur){
            if(tot>ans){
				ans=tot;
				return 1;
			}
            return 0;
        }
        for(int i=0;i<cur;i++){
			int u=Alt[tot][i],nxt=0;
            if(cur-i+tot<=ans)return 0;
            if(Max[u]+tot<=ans)return 0;
            for(int j=i+1;j<cur;j++)
                if(G[u][Alt[tot][j]])Alt[tot+1][nxt++]=Alt[tot][j];
            if(DFS(nxt,tot+1))return 1;
        }
        return 0;
    }
    int MaxClique(){
        ans=0,memset(Max,0,sizeof(Max));
        for(int i=n-1;i>=0;i--){
            int cur=0;
            for(int j=i+1;j<n;j++)
                if(G[i][j])Alt[1][cur++]=j;
            DFS(cur,1);
            Max[i]=ans;
        }
        return ans;
    }
}Group;

AC代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct MAX_CLIQUE{
    static const int N=110;
    bool G[N][N];
    int n,Max[N],Alt[N][N],ans;
    int y[N];
    int x[N];
    int *path,*res;
    bool DFS(int cur,int tot){
        if(!cur){
            if(tot>ans){
				ans=tot;
				swap(path,res);
				return 1;
			}
            return 0;
        }
        for(int i=0;i<cur;i++){
			int u=Alt[tot][i],nxt=0;
            if(cur-i+tot<=ans)return 0;
            if(Max[u]+tot<=ans)return 0;
            path[tot+1]=u;
            for(int j=i+1;j<cur;j++)
                if(G[u][Alt[tot][j]])Alt[tot+1][nxt++]=Alt[tot][j];
            if(DFS(nxt,tot+1))return 1;
        }
        return 0;
    }
    int MaxClique(){
        ans=0,memset(Max,0,sizeof(Max));
        path=x,res=y;
        for(int i=n-1;i>=0;i--){
            int cur=0;
            for(int j=i+1;j<n;j++)
                if(G[i][j])Alt[1][cur++]=j;
            path[1]=i;
            DFS(cur,1);
            Max[i]=ans;
        }
        return ans;
    }
}Group;    //模板
int main(){
	int m;
	scanf("%d",&m);
	while(m--){
		int k;
		int a,b;

		scanf("%d%d",&Group.n,&k);
		memset(Group.G,true,sizeof(Group.G));
		while(k--){
			scanf("%d%d",&a,&b);
			Group.G[a-1][b-1]=Group.G[b-1][a-1]=0;
		}

		int x=Group.MaxClique();

		printf("%d\n",x);
		for(int i=1;i<=x;i++){
			printf("%d ",Group.res[i]+1);
		}
		printf("\n");
	}
	return 0;
}