1142. Maximal Clique (25)

A clique is a subset of vertices of an undirected graph such that every two distinct vertices in the clique are adjacent. A maximal clique is a clique that cannot be extended by including one more adjacent vertex. (Quoted from https://en.wikipedia.org/wiki/Clique_(graph_theory))

Now it is your job to judge if a given subset of vertices can form a maximal clique.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers Nv (<= 200), the number of vertices in the graph, and Ne, the number of undirected edges. Then Ne lines follow, each gives a pair of vertices of an edge. The vertices are numbered from 1 to Nv.

After the graph, there is another positive integer M (<= 100). Then M lines of query follow, each first gives a positive number K (<= Nv), then followed by a sequence of K distinct vertices. All the numbers in a line are separated by a space.

Output Specification:

For each of the M queries, print in a line "Yes" if the given subset of vertices can form a maximal clique; or if it is a clique but not a maximal clique, print "Not Maximal"; or if it is not a clique at all, print "Not a Clique".

Sample Input:

8 10
5 6
7 8
6 4
3 6
4 5
2 3
8 2
2 7
5 3
3 4
6
4 5 4 3 6
3 2 8 7
2 2 3
1 1
3 4 3 6
3 3 2 1

Sample Output:

Yes
Yes
Yes
Yes
Not Maximal
Not a Clique


代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#define Max 100005
using namespace std;
int nv,ne,m,k;
char re[3][20] = {"Not a Clique","Not Maximal","Yes"};
int mp[201][201],vi[201],u[40001],v[40001],fir[40001],nex[40001],vis[201];
int check()
{
    for(int i = 0;i < k;i ++)///判断集合内任意两点是否连通
    {
        for(int j = i + 1;j < k;j ++)
        {
            if(!mp[vi[i]][vi[j]])return 0;
        }
    }
///满足clique
    for(int i = 1;i <= nv;i ++)///判断集合外是否存在一点与集合内点都连通
    {
        if(!vis[i])
        {
            int kk = fir[i],c = 0;
            while(kk != -1)
            {
                if(vis[v[kk]])c ++;
                if(c >= k)return 1;
                kk = nex[kk];
            }
        }
    }
///满足maximal
    return 2;
}
int main()
{
    scanf("%d%d",&nv,&ne);
    memset(fir,-1,sizeof(fir));
    for(int i = 0;i < ne;i ++)
    {
        scanf("%d%d",&u[i],&v[i]);
        if(u[i] == v[i])i --,ne --;
    }
    for(int i = 0;i < ne;i ++)
    {
        mp[u[i]][v[i]] = mp[v[i]][u[i]] = 1;
        u[i + ne] = v[i];
        v[i + ne] = u[i];
        nex[i] = fir[u[i]];
        fir[u[i]] = i;
        nex[i + ne] = fir[u[i + ne]];
        fir[u[i + ne]] = i + ne;
    }
    scanf("%d",&m);
    while(m --)
    {
        scanf("%d",&k);
        memset(vis,0,sizeof(vis));
        for(int i = 0;i < k;i ++)
        {
            scanf("%d",&vi[i]);
            vis[vi[i]] = 1;
        }
        puts(re[check()]);
    }
}