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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P2962 [USACO09NOV]Lights G

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P2962 [USACO09NOV]Lights G

直接上高斯消元。

考虑如何找到最优解，高斯消元后直接暴搜，对于自由元，枚举是否选择；否则如果不是自由元，增加答案。

查看代码 
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#include <cstdio>
#include <bitset>
using namespace std;
template <class Type>
void read(Type &x)
{
    char c;
    bool flag = false;
    while ((c = getchar()) < '0' || c > '9')
        c == '-' && (flag = true);
    x = c - '0';
    while ((c = getchar()) >= '0' && c <= '9')
        x = (x << 3) + (x << 1) + c - '0';
    flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
    x < 0 && (putchar('-'), x = ~x + 1);
    x > 9 && (write(x / 10), 0);
    putchar(x % 10 + '0');
}
const int N = 40;
bool vis[N];
int n, m, ans;
bitset<N> A[N];
void dfs (int x, int s)
{
    if (s >= ans)
        return;
    if (!x)
        return ans = s, void();
    if (A[x][x])
    {
        bool v = A[x][0];
        for (int i = x + 1; i <= n; i++)
            A[x][i] && (v ^= vis[i]);
        dfs(x - 1, s + v);
    }
    else
    {
        dfs(x - 1, s);
        vis[x] = true;
        dfs(x - 1, s + 1);
        vis[x] = false;
    }
}
void Gauss()
{
    for (int k = 1; k <= n; k++)
    {
        int t = k;
        while (t <= n && !A[t][k])
            t++;
        if (t > n) // No Solution
            continue;
        swap(A[t], A[k]);
        for (int i = 1; i <= n; i++)
            i ^ k && A[i][k] && (A[i] ^= A[k], 0);
    }
}
int main ()
{
    read(n), read(m);
    for (int a, b; m; m--)
    {
        read(a), read(b);
        A[a][b] = A[b][a] = 1;
    }
    for (int i = 1; i <= n; i++)
    {
        A[i][i] = 1;
        A[i][0] = 1;
    }
    Gauss();
    ans = n;
    dfs(n, 0);
    write(ans);
    return 0;
}

另外一种双向搜索的做法，将第一次搜索的中间状态塞进一个 map 里面，第二次再来查询。

查看代码 
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#include <cstdio>
#include <map>
using namespace std;
typedef long long L;
const int N = 35;
int n, m, ans = N;
L ed, op[N];
map <L, int> f;
void chkmin (int &x, int k) { if (k < x) x = k; }
void dfs (int x, int y, L s, int t)
{
	if (x == y)
	{
		if (y == n)
		{
			if (f.count(s))
				chkmin(ans, f[s] + t);
		}
		else
		{
			if (f.count(s)) chkmin(f[s], t);
			else f[s] = t;
		}
		return;
	}
	dfs(x + 1, y, s ^ op[x], t + 1);
	dfs(x + 1, y, s, t);
}
int main()
{
	scanf("%d%d", &n, &m);
	ed = (1ll << n) - 1;
	for (int i = 0; i < n; ++i)
		op[i] |= 1ll << i;
	for (int i = 1, a, b; i <= m; ++i)
	{
		scanf("%d%d", &a, &b);
		--a, --b;
		op[a] |= 1ll << b;
		op[b] |= 1ll << a;
	}
//	for (int i = 0; i < n; ++i)
//		printf("%lld\n", op[i]);
//	return 0;
	dfs(0, n >> 1, 0, 0);
	dfs(n >> 1, n, ed, 0);
	printf("%d", ans);
 	return 0;
}
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