PAT 甲级 1142 Maximal Clique
https://pintia.cn/problem-sets/994805342720868352/problems/994805343979159552
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
代码:
#include <bits/stdc++.h>
using namespace std; int N, M, K;
int mp[220][220];
int vis[220]; int main() {
scanf("%d%d", &N, &M);
memset(mp, 0, sizeof(mp));
while(M --) {
int a, b;
scanf("%d%d", &a, &b);
mp[a][b] = mp[b][a] = 1;
}
scanf("%d", &K);
while(K --) {
int T;
vector<int> v;
memset(vis, 0, sizeof(vis));
scanf("%d", &T);
v.resize(T);
for(int i = 0; i < T; i ++) {
scanf("%d", &v[i]);
vis[v[i]] = 1;
} bool isclique = true;
for(int i = 0; i < T - 1; i ++) {
for(int j = i + 1; j < T; j ++) {
if(mp[v[i]][v[j]] == 0) {
isclique = false;
printf("Not a Clique\n");
break;
}
}
if(!isclique) break;
} if(!isclique) continue; bool ismax = true;
for(int i = 1; i <= N; i ++) {
if(vis[i] == 0) {
for(int j = 0; j < T; j ++) {
if(mp[v[j]][i] == 0) break;
if(j == T - 1) ismax = false;
}
}
if(!ismax) {
printf("Not Maximal\n");
break;
}
}
if(ismax) printf("Yes\n");
}
return 0;
}
简单图论
FHFHFH
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