题面
Problem Description
Sociologists are interested in the phenomenon of "friendship". To study this property, they analyze various groups of people. For each two persons in such a group they determine whether they are friends (it is assumed that this relation is symmetric). The sociologists are mostly interested in the sets of friends. The set S of people is the set of friends if every two persons in S are friends. However, studying the sets of friends turns out to be quite complicated, since there are too many such sets. Therefore, they concentrate just on the maximal sets of friends. A set of friends S is maximal if every person that does not belong to S is not a friend with someone in S.
Your task is to determine the number of maximal sets of friends in each group. In case this number exceeds 1 000, you just need to report this -- such a group is too complicated to study.
Input
The input consists of several instances, separated by single empty lines.
The first line of each instance consists of two integers 1 ≤ n ≤ 128 and m -- number of persons in the group and number of friendship relations. Each of m following lines consists of two integers ai and bi (1 ≤ ai, bi ≤ n). This means that persons ai and bi (ai ≠ bi) are friends. Each such relationship is described at most once.
Output
The output for each instance consists of a single line containing the number of maximal sets of friends in the described group, or string "Too many maximal sets of friends." in case this number is greater than 1 000.
Sample Input
5 4
1 2
3 4
2 3
4 5
Sample Output
4
思路
极大团裸题。用BK算法解决。
代码实现
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=0;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005
#define fi first
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
char ch=getchar(); int x=0, f=1;
while(ch<'0'||ch>'9') {
if(ch=='-') f = -1;
ch=getchar();
}
while('0'<=ch&&ch<='9') {
x=x*10+ch-'0';
ch=getchar();
} return x*f;
}
const int maxn=200;
int g[maxn][maxn];
int all[maxn][maxn],some[maxn][maxn],none[maxn][maxn];
int ans,n,m,s;
void dfs (int n,int an,int sn,int nn) {
if (!sn&&!nn) ans++;
if (ans>1000) return ;
int key=some[n][1];
rep (j,1,sn) {
int v=some[n][j],tsn=0,tnn=0;
if (g[key][v]) continue;
rep (i,1,an) all[n+1][i]=all[n][i]; all[n+1][an+1]=v;
rep (i,1,sn) if (g[v][some[n][i]]) some[n+1][++tsn]=some[n][i];
rep (i,1,nn) if (g[v][none[n][i]]) none[n+1][++tnn]=none[n][i];
dfs (n+1,an+1,tsn,tnn);
some[n][j]=0,none[n][++nn]=v;
}
}
int main () {
while (cin>>n>>m) {
MT (g,0);
ans=0;
rep (i,1,m) {
int x,y;
x=read (),y=read ();
g[x][y]=g[y][x]=1;
}
rep (i,1,n) some[1][i]=i;
dfs (1,0,n,0);
if (ans>1000) cout<<"Too many maximal sets of friends."<<endl;
else cout<<ans<<endl;
}
return 0;
}