POJ - 1419 （最大独立集 最大团问题）

Graph Coloring

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5414		Accepted: 2526		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

Source

Southwestern European Regional Contest 1995

解题思路：求解无向图的最大独立集。无向图最大独立集=该无向图的补图的最大团。所以用最大团求解。关于最大团网上有很多介绍了。直接暴力搜索解即可。详见代码

//最大团做法

#include <iostream>
#include <string>
#include <algorithm>
#include <map>
#include <memory.h>
#include <stack>

using namespace std;


int G[105][105];//图
int n,m;
int inset[105];//当前在团中的点
int chosen[105];//最大团的点集合
int nownum;//当前最大团点的数量
int bestnum;//最大团点的数量
void dfs(int i){
    //如果所有节点都被深搜了
    if(i>n){
        memcpy(chosen,inset,sizeof(chosen));//更新最大团的点和数量
        bestnum=nownum;
        return;
    }
    bool flag=1;//当前点是否与当前最大团中的点都相连
    for(int j=0;j<nownum;j++){
        if( G[i][inset[j]]==0){
            flag=false;
            break;
        }
    }
    //是的话，可以加入到最大团中
    if(flag){
        inset[nownum++]=i;
        dfs(i+1);
        nownum--;
    }
    
    //否则不加人，和一个小小的剪枝
    if(nownum+(n-i)>bestnum)
        dfs(i+1);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        int a,b;
        memset(inset,0,sizeof(inset));
        memset(G,1,sizeof(G));//求补图
        nownum=0;
        bestnum=0;
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            G[a][b]=0;//补图
            G[b][a]=0;
        }
        dfs(1);//从第一个节点开始深搜

        printf("%d\n",bestnum);
        for(int i=0;i<bestnum;i++){
            if(i!=bestnum-1)
                printf("%d ",chosen[i]);
            else
                printf("%d\n",chosen[i]);
        }


    }

    return 0;

}

//最大独立集做法（实际上就是把0改成1,1改成0）

#include <iostream>
#include <string>
#include <algorithm>
#include <map>
#include <memory.h>
#include <stack>

using namespace std;


int G[105][105];//图
int n,m;
int inset[105];//当前在独立集中的点
int chosen[105];//最大独立集的点集合
int nownum;//当前最大独立集的点的数量
int bestnum;//最大独立集点的数量
void dfs(int i){
    //如果所有节点都被深搜了
    if(i>n){
        memcpy(chosen,inset,sizeof(chosen));//更新最大独立集的点和数量
        bestnum=nownum;
        return;
    }
    bool flag=1;//当前点是否有与当前最大独立集中的点相连
    for(int j=0;j<nownum;j++){
        if( G[i][inset[j]]==1){
            flag=false;
            break;
        }
    }
    //没有的话，可以加入到最大独立集中
    if(flag){
        inset[nownum++]=i;
        dfs(i+1);
        nownum--;
    }

    
    if(nownum+(n-i)>bestnum)
        dfs(i+1);
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        int a,b;
        memset(inset,0,sizeof(inset));
        memset(G,0,sizeof(G));
        nownum=0;
        bestnum=0;
        for(int i=0;i<m;i++){
            scanf("%d%d",&a,&b);
            G[a][b]=1;
            G[b][a]=1;
        }
        dfs(1);//从第一个节点开始深搜

        printf("%d\n",bestnum);
        for(int i=0;i<bestnum;i++){
            if(i!=bestnum-1)
                printf("%d ",chosen[i]);
            else
                printf("%d\n",chosen[i]);
        }


    }

    return 0;

}