题目如下
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
Clique是一个点集,其中所有顶点两两相连,Maximal Clique就是说这个Clique使得在给定的图中没有更多的点能加入这个Clique形成新的Clique。
不难,用矩阵保存边集,设置两个标志量,标识这个集合是不是Clique、这个Clique是不是Maximal Clique。缓存每一个样例,先检查是不是两两相连,如果是,检查图中剩下的顶点中是否存在顶点v和当前Clique中所有顶点相连。顶点是1-n排列的,省了去一点麻烦。
代码如下
#include <iostream>
#include <vector>
using namespace std;
int e[201][201];
int main() {
int nv, ne, m, a, b, k;
scanf("%d %d", &nv, &ne);
for (int i = 0; i < ne; i++) {
scanf("%d %d", &a, &b);
e[a][b] = e[b][a] = 1;
}
scanf("%d", &m);
for (int i = 0; i < m; i++) {
scanf("%d", &k);
vector<int> v(k);
int check[201] = { 0 }, isClique = 1, isMaximal = 1;
for (int j = 0; j < k; j++) {
scanf("%d", &v[j]);
check[v[j]] = 1;
}//保存样例并记录在check中
for (int j = 0; j < k; j++) {
if (isClique == 0) break;
for (int l = j + 1; l < k; l++) {
if (e[v[j]][v[l]] == 0) {
isClique = 0;
printf("Not a Clique\n");
break;
}
}
}//判断Clique,不是就continue
if (isClique == 0) continue;
for (int j = 1; j <= nv; j++) {
if (check[j] == 0) {
for (int l = 0; l < k; l++) {
if (e[v[l]][j] == 0) break;
if (l == k - 1) isMaximal = 0;//查到最后还没有break,没有符合条件的点
}//注意不要习惯性写成一旦==1就把标志变成0,我这样错过。
}
if (isMaximal == 0) {
printf("Not Maximal\n");
break;
}
}
if (isMaximal == 1) {
printf("Yes\n");
}
}
return 0;
}