HDU-5952 爆搜DFS

HDU5952

题目大意

给一个\(n\)个点\(m\)条边的无向图，要求你选择其中\(s\)个点，使齐构成完全图，问满足要求的选法有几种。

\(N≤100\)

\(M≤1000\)

\(2≤S≤10\)

思路

数据范围那么小，显然是爆搜。

我们可以通过爆搜大小为s的团来得到答案，但是，普通爆搜肯定会TLE的

因为其中搜索了大量重复的团，例如：1-3-4-5其实和1-4-5-3是一样的

那么我们可以在搜索中保证团的大小一定是从小到大的，例如1-2-3-4 1-3-4-5

这样就避免了搜索重复的团

Code

#include<string>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<vector>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define LL long long
#define MOD 998244353
#define PI 3.1415926535898
#define INF 0x3f3f3f3f
#define MAXN 300
const double EPS = 1e-8;
LL read()
{
	LL x = 0, w = 1;
	char ch = 0;
	while (ch < '0' || ch>'9')
	{
		if (ch == '-')
		{
			w = -1;
		}
		ch = getchar();
	}
	while (ch >= '0' && ch <= '9')
	{
		x = x * 10 + ch - '0';
		ch = getchar();
	}
	return w * x;
}
LL t, n, s, m, ui, vi, wi, ans;
LL mp[MAXN][MAXN], a[MAXN], as[MAXN][MAXN];
bool v[MAXN][MAXN], vis[MAXN];
LL head[MAXN], cnt = 0;
struct node
{
	LL to, w, next;
};
node edge[20005];
inline void add(LL x, LL y, LL w)
{
	edge[++cnt].to = y;
	edge[cnt].next = head[x];
	head[x] = cnt;
	edge[cnt].w = w;
}
inline void init()
{
	memset(edge, 0, sizeof(edge));
	memset(a, 0, sizeof(a));
	memset(mp, 0, sizeof(mp));
	memset(head, 0, sizeof(head));
	cnt = 0;
	ans = 0;
}
LL dfs(LL x, LL cn)
{
	LL sum = 0;
	if (cn == s)
	{
		ans++;
		v[x][cn] = 1;
		as[x][cn] = 1;
		return 1;
	}
	a[cn] = x;
	for (register LL i = head[x]; i; i = edge[i].next)
	{
		bool f = 1;
		for (register LL j = 1; j <= cn; j++)
		{
			if (mp[a[j]][edge[i].to] == 0)
			{
				f = 0;
				break;
			}
		}
		if (f)
		{
			dfs(edge[i].to, cn + 1);
		}

	}
	return sum;
}
int main()
{
	t = read();
	while (t--)
	{
		init();
		n = read();
		m = read();
		s = read();
		for (register LL i = 1; i <= m; i++)
		{
			ui = read();
			vi = read();
			//wi = read();
			if (ui > vi)
				swap(ui, vi);
			add(ui, vi, 1);
			mp[ui][vi] = 1;
			mp[vi][ui] = 1;
		}
		for (register LL i = 1; i <= n; i++)
		{
			dfs(i, 1);
		}
		cout << ans << endl;
	}
	return 0;
}