P4906 小奔关闹钟

${color{Cyan}{>>Question}}$

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

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std; 

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//-------------------------------------------------------

const int N = 21;

int n,m,ans = 1e9+7;
int op[N],a[N][N];

void init() {
    int i,j,k;
    for(i = 0;i < n; ++i) {
        op[i] ^= (1<<i);
        for(j = 0;j < n; ++j) {
            if(a[i][j] && j != i) {
                op[i] ^= (1<<j);
                for(k = 0;k < n; ++k) {
                    if(a[j][k] && k != j)
                        op[i] ^= (1<<k);
                }
            }
        }
    }
}

void dfs(int k,int s,int cnt) {
    if(cnt >= ans) return;
    if(k == n) {if(!s) ans = cnt; return;}
    dfs(k+1,s^op[k],cnt+1);
    dfs(k+1,s,cnt);
}

int main() {
    int i,j,k,x;
    in(n);
    for(i = 0;i < n; ++i) {
        in(m);
        for(j = 1;j <= m; ++j) {
            in(x); --x; a[i][x] = 1;
        }
    }
    init();
    dfs(0,(1<<n)-1,0);
    if(ans == 1e9+7) cout << "Change an alarm clock，please!";
    else out(ans);
    return 0;
}