P4906 小奔关闹钟

Posted 2022-07-14 mzg1805

$\colorCyan>>Question$

一道水题

注意题中只要求扩展两层以内的开关

所以预处理出按了开关的影响,利用异或的性质就能很好地解决

然后$dfs$时枚举每个开关按或不按(因为即使再按也是相同的情况)

就没了$?$ 没了

代码如下

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cstring>
 5 #define ll long long
 6 using namespace std; 
 7 
 8 template <typename T> void in(T &x) 
 9     x = 0; T f = 1; char ch = getchar();
10     while(!isdigit(ch)) if(ch == ‘-‘) f = -1; ch = getchar();
11     while( isdigit(ch)) x = 10 * x + ch - 48; ch = getchar();
12     x *= f;
13 
14 
15 template <typename T> void out(T x) 
16     if(x < 0) x = -x , putchar(‘-‘);
17     if(x > 9) out(x/10);
18     putchar(x%10 + 48);
19 
20 //-------------------------------------------------------
21 
22 const int N = 21;
23 
24 int n,m,ans = 1e9+7;
25 int op[N],a[N][N];
26 
27 void init() 
28     int i,j,k;
29     for(i = 0;i < n; ++i) 
30         op[i] ^= (1<<i);
31         for(j = 0;j < n; ++j) 
32             if(a[i][j] && j != i) 
33                 op[i] ^= (1<<j);
34                 for(k = 0;k < n; ++k) 
35                     if(a[j][k] && k != j)
36                         op[i] ^= (1<<k);
37                 
38             
39         
40     
41 
42 
43 void dfs(int k,int s,int cnt) 
44     if(cnt >= ans) return;
45     if(k == n) if(!s) ans = cnt; return;
46     dfs(k+1,s^op[k],cnt+1);
47     dfs(k+1,s,cnt);
48 
49 
50 int main() 
51     int i,j,k,x;
52     in(n);
53     for(i = 0;i < n; ++i) 
54         in(m);
55         for(j = 1;j <= m; ++j) 
56             in(x); --x; a[i][x] = 1;
57         
58     
59     init();
60     dfs(0,(1<<n)-1,0);
61     if(ans == 1e9+7) cout << "Change an alarm clock，please!";
62     else out(ans);
63     return 0;
64