poj 1419 最大团

Graph Coloring

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3346		Accepted: 1491		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

给定一个图，给这个图的点着色，只有黑白两种颜色，并且相邻两点不能都着色为黑，问最多有几个点能被染成黑色。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
#define maxm 105
int n,m,ans,cnt;//ans存储最后的点数
int map[maxm][maxm],vis[maxm],opt[maxm];//map存图，opt标记最大团中的点
void dfs(int t)
{
    if(t>n)
    {
        ans=cnt;
        for(int i=1;i<=n;i++)
        opt[i]=vis[i];
        return ;
    }
    int flag=0;
    for(int j=1;j<t;j++)
    if(vis[j]&&map[j][t])
    {
        flag=1;
        break;
    }
    if(!flag)
    {
        vis[t]=1;
        cnt++;
        dfs(t+1);
        cnt--;
    }
    if(cnt+n-t>ans)//最大团中的点数加上没有被访问过的点数
                   //如果大于最大团点数，可进行剪枝
    {
        vis[t]=0;
        dfs(t+1);
    }
}
int main()
{
    int cas,a,b;
    cin>>cas;
    while(cas--)
    {
        memset(map,0,sizeof(map));
        ans=cnt=0;
        cin>>n>>m;
        for(int i=0;i<m;i++)
        {
            cin>>a>>b;
            map[a][b]=map[b][a]=1;
        }
        dfs(1);
        cout<<ans<<endl;
        for(int i=1;i<=n;i++)
        if(opt[i])
        cout<<i<<" ";
        cout<<endl;
    }
    return 0;
}