Problem #
Source: [PAT 1142]
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 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 1Sample Output: #
Yes
Yes
Yes
Yes
Not Maximal
Not a CliqueSolution #
- 题意 给你一张图，再给你一串序列，你判断是否是最极大团
- 思路 判断序列内是否两两可通；判断图内是否存在非团内点两两可通
Code #
#include <iostream>
#include <algorithm>
#include <set>
#define maxsize 202
using namespace std;
int n, m, a, b, matrx[maxsize][maxsize] = {0}, list[maxsize];
set<int> lists;
bool vis[maxsize];
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
cin >> n >> m;
for (int i = 0; i < m; i++)
{
cin >> a >> b;
matrx[a][b] = matrx[b][a] = 1;
}
cin >> m;
for (int i = 0; i < m; i++)
{
cin >> a;
lists.clear();
for (int j = 0; j < a; j++)
{
cin >> list[j];
lists.insert(list[j]);
}
bool flag = true;
for (int j = 0; j < a && flag; j++)
{
for (int z = 0; z < a && flag; z++)
{
if (matrx[list[j]][list[z]] != 1 && j != z)
flag = false;
}
}
if (!flag)
{
cout << "Not a Clique" << endl;
}
else
{
int s;
for (s = 1; s <= n; s++)
{
flag = true;
if (lists.find(s) == lists.end())
{
for (int z = 0; z < a && flag; z++)
{
if (matrx[s][list[z]] != 1)
flag = false;
}
if (flag)
{
cout << "Not Maximal" << endl;
break;
}
}
}
if (s == n + 1)
{
cout << "Yes" << endl;
}
}
}
return 0;
}
