Problem Description:
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 N v N_v Nv ( ≤ 200 \leq 200 ≤200), the number of vertices in the graph, and N e N_e Ne, the number of undirected edges. Then N e N_e Ne lines follow, each gives a pair of vertices of an edge. The vertices are numbered from 1 to N v N_v Nv.
After the graph, there is another positive integer M M M ( ≤ 100 \leq 100 ≤100). Then M M M lines of query follow, each first gives a positive number K K K ( ≤ N v \leq N_v ≤Nv), then followed by a sequence of K K K distinct vertices. All the numbers in a line are separated by a space.
Output Specification:
For each of the M M 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
Problem Analysis:
题目中每组询问给定一个顶点集合,需要我们判断该集合中的点和边是否成团,如果成团,是否是最大团。
对于给定的一个顶点集合,我们可以判断任何两个点之间是否存在边,如果是,则是一个团。
bool check_clique(int cnt)
{
for (int i = 1; i <= cnt; i ++ )
for (int j = i + 1; j <= cnt; j ++ )
if (!g[vers[i]][vers[j]])
return false;
return true;
}
对于判断是否是最大团,在已经确认是团的基础上,我们可以遍历所有不在这个团的所有点,一旦存在这样一个点,这个点与团内所有点之间都存在边,那么就说明这个点可以加入这个团,那么就证明这个团不是一个最大团:
bool check_maximal(int cnt)
{
memset(st, 0, sizeof st);
for (int i = 1; i <= cnt; i ++ )
st[vers[i]] = true;
bool success = false;
for (int i = 1; i <= n; i ++ )
if (!st[i]) // 判断不在原集合的所有点
{
// 我们来判断除了给定集合,还有没有其他的点能够参与进团
int e = 0;
for (int j = 1; j <= cnt; j ++ )
if (g[i][vers[j]]) ++ e;
if (e == cnt) return false;
}
return true;
}
Code
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 210;
int n, m;
bool g[N][N];
int vers[N];
bool st[N];
bool check_clique(int cnt)
{
for (int i = 1; i <= cnt; i ++ )
for (int j = i + 1; j <= cnt; j ++ )
if (!g[vers[i]][vers[j]])
return false;
return true;
}
bool check_maximal(int cnt)
{
memset(st, 0, sizeof st);
for (int i = 1; i <= cnt; i ++ )
st[vers[i]] = true;
bool success = false;
for (int i = 1; i <= n; i ++ )
if (!st[i]) // 判断不在原集合的所有点
{
// 我们来判断除了给定集合,还有没有其他的点能够参与进团
int e = 0;
for (int j = 1; j <= cnt; j ++ )
if (g[i][vers[j]]) ++ e;
if (e == cnt) return false;
}
return true;
}
int main()
{
cin >> n >> m;
while (m -- )
{
int a, b;
cin >> a >> b;
g[a][b] = g[b][a] = true;
}
int k;
cin >> k;
while (k -- )
{
int cnt;
cin >> cnt;
for (int i = 1; i <= cnt; i ++ ) scanf("%d", &vers[i]);
if (check_clique(cnt))
{
if (check_maximal(cnt)) puts("Yes");
else puts("Not Maximal");
}
else puts("Not a Clique");
}
return 0;
}