POJ1419 Graph Coloring（最大独立集）（最大团）

                                                           Graph Coloring

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 4926		Accepted: 2289		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
【分析】此题就是求最大独立集。二部图的最大独立集==顶点数-匹配数，普通图的最大独立集==补图的最大团，且此题需要输出路径，所以用最大团的模板算法较好。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <queue>
#include <vector>
#define inf 0x7fffffff
#define met(a,b) memset(a,b,sizeof a)
typedef long long ll;
using namespace std;
const int N = 105;
const int M = 25005;
bool w[N][N];
bool use[N]; //进入团的标号
bool bestx[N];
int cn,bestn,p,e;

void dfs(int x) {
    bool flag;
    if(x>p) {
        bestn=cn; //cn的值是递增的
        for( int i=1;i<=p; i++) //赋值给另外一个数组，
            bestx[i]=use[i];
        return ;
    }
    flag=true;
    for( int i=1; i<x; i++)
        if(use[i]&&!w[i][x]) {
            flag=false;
            break;
        }
    if(flag) {
        cn++;
        use[x]=true;
        dfs(x+1);
        cn--;
        use[x]=false;//回溯
    }
    if(cn+p-x>bestn) { //剪枝
        dfs(x+1);
    }
}

int main() {
    int num,u,v;
    scanf("%d",&num);
    while(num--) {
        memset(w,true,sizeof(w));
        memset(use,false,sizeof(use));
        memset(bestx,false,sizeof(bestx));
        scanf("%d%d",&p,&e);
        for(int i=0; i<e; i++) {
            scanf("%d%d",&u,&v);
            w[u][v]=false;
            w[v][u]=false;
        }
        cn=bestn=0;
        dfs(1);
        printf("%d
",bestn);
        for (int i=1; i<=p; i++)if(bestx[i])printf("%d ",i);printf("
");
    }
    return 0;
}