目录 题目链接 题解 题目链接 luogu P2962 [USACO09NOV]灯Lights 题解 可以折半搜索 map合并 复杂度 2^(n / 2)*logn 高斯消元后得到每个点的翻转状态 爆搜自由元得到最优翻转状态 // luogu-judger-enable-o2 #include<map> #include<queue> #include<cstdio> #include<cstring> #include<algorithm> u…

\(\color{#0066ff}{题目描述}\) 贝希和她的闺密们在她们的牛棚中玩游戏.但是天不从人愿,突然,牛棚的电源跳闸了,所有的灯都被关闭了.贝希是一个很胆小的女生,在伸手不见拇指的无尽的黑暗中,她感到惊恐,痛苦与绝望.她希望您能够帮帮她,把所有的灯都给重新开起来!她才能继续快乐地跟她的闺密们继续玩游戏! 牛棚中一共有N(1 <= N <= 35)盏灯,编号为1到N.这些灯被置于一个非常複杂的网络之中.有M(1 <= M <= 595)条很神奇的无向边,每条边连接两盏灯.…

题目描述 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…

题目描述 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…

题目描述 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…

Description 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.…

贝希和她的闺密们在她们的牛棚中玩游戏.但是天不从人愿,突然,牛棚的电源跳闸了,所有的灯都被关闭了.贝希是一个很胆小的女生,在伸手不见拇指的无尽的黑暗中,她感到惊恐,痛苦与绝望.她希望您能够帮帮她,把所有的灯都给重新开起来!她才能继续快乐地跟她的闺密们继续玩游戏! 牛棚中一共有N(1 <= N <= 35)盏灯,编号为1到N.这些灯被置于一个非常複杂的网络之中.有M(1 <= M <= 595)条很神奇的无向边,每条边连接两盏灯. 每盏灯上面都带有一个开关.当按下某一盏灯的开关的时候…

传送门 先进行高斯消元 因为要求最少的开关次数,那么: 对于关键元,我们可以通过带入消元求出, 对于自由元,我们暴力枚举,进行dfs,因为只有开关两种状态,0或1 #include <cmath> #include <cstdio> #include <iostream> #define N 40 using namespace std; int n, m, sum, mn = ~(1 << 31); int a[N][N], ans[N]; inline…

题目链接 本意是想学高斯消元,然后一顿乱搞之后学到了一个神奇的搜索方式叫做折半搜索. qwq 就是我先dfs前二分之n个点,然后再dfs后二分之n个点. 然后我dfs后二分之n个点的时候判断一下第一次dfs有没有搜到互补的状态(就是当前状态能不能跟之前搜到过的一个状态异或起来变成(1<<n)-1,即所有灯都开着) 如果能就看看能不能更新答案. 然后快的飞起qwq #include<cstdio> #include<algorithm> #include<cstri…

复杂度分析 假设本来是n层,本来复杂度是O(2^n),如果meet in middle那就是n/2层,那复杂度变为O( 2^(n/2) ),跟原来的复杂度相比就相当于开了个方 比如如果n=40那爆搜2^40肯定T飞,那用meet in middle的话就是2^20就可做了. 洛谷P2962 [USACO09NOV]灯Lights 灯只有35个,用二进制可以表示所有灯的状态,于是考虑搜索 1表示该灯是亮的,0表示是灭的 把某一个等和与他相邻的灯的位都置1表示该灯位置的开关,用 li 数组表示,按下…