1142 Maximal Clique （25 分）

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 ))

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:

8 105 67 86 43 64 52 38 22 75 33 464 5 4 3 63 2 8 72 2 31 13 4 3 63 3 2 1

Sample Output:

YesYesYesYesNot MaximalNot a Clique

 题目大意：clique是一个点集，在一个无向图中，这个点集中任意两个不同的点之间都是相连的。maximal clique是一个clique，这个clique不可以再加入任何一个新的结点构成新的clique。

输入是有n条边，给出每条边的两端节点，并且后面给出m个查询，查询是点集。

 //这个题目大意看了好几遍没看懂，这个clique也不是环，比如对第三个查询2 3，输出Yes，说明不是环了。

//思考了一下发现不太会，怎么去确定这个是maximal的呢？怎么去扩展判断呢？不会。

代码转自：https://www.liuchuo.net/archives/4614

#include 
     
      #include 
      
       #include
       
        using namespace std;int e[210][210];int main() {    int nv, ne, m, ta, tb, k;    scanf("%d %d", &nv, &ne);    for (int i = 0; i < ne; i++) {        scanf("%d %d", &ta, &tb);        e[ta][tb] = e[tb][ta] = 1;//存储到邻接矩阵中，有边是1.    }    scanf("%d", &m);    for (int i = 0; i < m; i++) {        scanf("%d", &k);        vector
        
          v(k);        int hash[210] = {
   0}, isclique = 1, isMaximal = 1;        for (int j = 0; j < k; j++) {            scanf("%d", &v[j]);            hash[v[j]] = 1;//使用hash数组存储        }        for (int j = 0; j < k; j++) {
   //这里判断是否是一个click            if (isclique == 0) break;//跳出两层循环            for (int l = j + 1; l < k; l++) {                if (e[v[j]][v[l]] == 0) {                    isclique = 0;                    printf("Not a Clique\n");                    break;                }            }        }        if (isclique == 0) continue;//不进行下面的操作。        for (int j = 1; j <= nv; j++) {            if (hash[j] == 0) {
   //挨个判断其他所有的点，判断每一个点。                for (int l = 0; l < k; l++) {
   //和当前检测中所有的点进行判断。                    if (e[v[l]][j] == 0) break;//如果这个点不是的话，接着判断其他点                    if (l == k - 1) isMaximal = 0;                }            }            if (isMaximal == 0) {                printf("Not Maximal\n");                break;            }        }        if (isMaximal == 1) printf("Yes\n");    }    return 0;}
        
       
      
     

 

//柳神真厉害。

1.使用邻接矩阵存储图，右边标记为1.

2.对于输入的使用hash数组来标记，向量来存储

3.对图中所有剩下的点一一与当前检测中的进行判断。

 //学习了！