PAT甲级2017冬7-3 1142 Maximal Clique (25分)

客官°小女子只卖身不卖艺 2023-02-25 10:08 127阅读 0赞

[算法笔记总目录][Link 1]  
[关键英语单词解释][Link 2]  
本题代码与[PAT甲级2019冬7-3 Summit（1166）][PAT_2019_7-3 Summit_1166]除输出不一样外，其他不变。  
[1142 Maximal Clique (25分)][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

#include<bits/stdc++.h>
    using namespace std;
    int e[205][205];
    int a[205];
    int v[205];
    int n,m;
    int main(){ 
        cin>>n>>m;
        memset(e,0,sizeof(e));
        for(int i=1;i<=m;i++){ 
            int u,v;
            cin>>u>>v;
            e[u][v]=e[v][u]=1;//邻接矩阵存储 
        }
        int k;
        cin>>k;
        int num;
        for(int i=1;i<=k;i++){ //k个询问 
            cin>>num;
            memset(v,0,sizeof(v));//v来标记所询问的num个点 
            for(int j=1;j<=num;j++) { 
                cin>>a[j];
                v[a[j]]=1;//做上标记 
            }
            int f=1;//是不是两两相连 
            for(int j=1;j<=num;j++){ 
                for(int p=j+1;p<=num;p++){ 
                    if(e[a[j]][a[p]]!=1) f=0;
                    break;
                }
            }
            if(!f) cout<<"Not a Clique";
            else{ //如果是两两相连 
                int ans=-1;//是否存在 
                for(int j=1;j<=n;j++){ //查询存不存在一个点与这num个点都相连 
                    if(v[j]) continue;//本身是num个点里的不算 
                    int ff=1; 
                    for(int p=1;p<=num;p++){ 
                        if(e[a[p]][j]!=1){ 
                            ff=0;
                            break;    
                        }
                    }
                    if(ff) { //满足与num中的每个点都相连 
                        ans=j;//存在 
                        break;
                    }
                }
                if(ans!=-1){ //存在 
                    cout<<"Not Maximal";
                }else{ //不存在 
                    cout<<"Yes"; 
                }
            }
            if(i!=k) cout<<endl;//行末无空行 
        }
        return 0; 
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]

[Link 1]: https://blog.csdn.net/qq_21480607/article/details/106660085
[Link 2]: https://blog.csdn.net/qq_21480607/article/details/106987591
[PAT_2019_7-3 Summit_1166]: https://blog.csdn.net/qq_21480607/article/details/106896828
[1142 Maximal Clique _25]: https://pintia.cn/problem-sets/994805342720868352/problems/994805343979159552
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]: /images/20230209/af904bb749b94e9a8e025b14bcff2979.png