题意：

有 n盏灯，每盏灯与若干盏灯相连，每盏灯上都有一个开关，如果按下一盏灯上的开关，这盏灯以及与之相连的所有灯的开关状态都会改变。一开始所有灯都是关着的，你需要将所有灯打开，求最小的按开关次数。(1<=n<=35)。

题目：

Bessie and the cows were playing games in the barn, but the power was reset and the lights were all turned off. Help the cows get all the lights back on so they can resume their games.

The N (1 <= N <= 35) lights conveniently numbered 1…N and their switches are arranged in a complex network with M (1 <= M <= 595) clever connection between pairs of lights (see below).

Each light has a switch that, when toggled, causes that light – and all of the lights that are connected to it – to change their states (from on to off, or off to on).

Find the minimum number of switches that need to be toggled in order to turn all the lights back on.

It’s guaranteed that there is at least one way to toggle the switches so all lights are back on.

Input format

Line 1: Two space-separated integers: N and M.
Lines 2…M+1: Each line contains two space-separated integers representing two lights that are connected. No pair will be repeated.

Output format

Line 1: A single integer representing the minimum number of switches that need to be flipped in order to turn on all the lights.

Explanation

There are 5 lights. Lights 1, 4, and 5 are each connected to both lights 2 and 3.

Toggle the switches on lights 1, 4, and 5.

输入

5 6
1 2
1 3
4 2
3 4
2 5
5 3

输出

3

分析：

1.如果这道题暴力 DFS 找开关灯的状态，时间复杂度就是$O(2^n)$ , 显然超时。不过，如果我们用 meet in middle 的话，时间复杂度可以优化至$O(n{2}^{\frac{n}{2}})$ 。
2.meet in middle 就是让我们先找一半的状态，也就是找出只使用编号为 1到 mid的开关能够到达的状态，再找出只使用另一半开关能到达的状态。
3.如果前半段和后半段开启的灯互补，将这两段合并起来就得到了一种将所有灯打开的方案。
4.具体实现时，可以把前半段的状态以及达到每种状态的最少按开关次数存储在 map 里面，搜索后半段时，每搜出一种方案，就把它与互补的第一段方案合并来更新答案。

AC代码：

#include <algorithm>
#include <cstdio>
#include <iostream>
#include <map>using namespace std;typedef long long ll;int n, m, ans = 0x7fffffff;
map<ll, int> f;
ll a[40];
int main()
{cin >> n >> m;for (int i = 0; i < n; ++i)a[i] = (1ll << i);for (int i = 1; i <= m; ++i){int u, v;cin >> u >> v;--u;--v;a[u] |= (1ll << v);a[v] |= (1ll << u);}for (int i = 0; i < (1 << (n / 2)); ++i){ll t = 0;int cnt = 0;for (int j = 0; j < n / 2; ++j){if ((i >> j) & 1){t ^= a[j];++cnt;}}if (!f.count(t))f[t] = cnt;elsef[t] = min(f[t], cnt);}for (int i = 0; i < (1 << (n - n / 2)); ++i){ll t = 0;int cnt = 0;for (int j = 0; j < (n - n / 2); ++j){if ((i >> j) & 1){t ^= a[n / 2 + j];++cnt;}}if (f.count(((1ll << n) - 1) ^ t))ans = min(ans, cnt + f[((1ll << n) - 1) ^ t]);}cout << ans;return 0;
}