费解的开关 Page15 优化枚举

如果暴力枚举复杂度是 $O(2^{32})$

根据性质可以先确定第一层，层层递推过去，复杂度被优化到了 $O(2^5×5^2)$

代码：

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2e6+7;
const int INF=1e9;
const ll INFF=1e18;
int a[6][6],aa[6][6],ans,T;
int x[6]={0,0,1,-1,0};
int y[6]={1,-1,0,0,0};
void click(int i,int j)
{
    rep(k,0,4)
    {
        int xx=i+x[k];
        int yy=j+y[k];
        if (xx<1||xx>5||yy<1||yy>5)continue;
        a[xx][yy]=1-a[xx][yy];
    }
}
void init()
{
    rep(i,1,5)rep(j,1,5)a[i][j]=aa[i][j];
}
void solve(int x)
{
    int num=0;
    rep(k,0,4)
    {
        if ((x>>k)&1)
        {
            click(1,k+1);
            num++;
        }
    }
    rep(i,1,4)
    {
        rep(j,1,5)
        {
            if (a[i][j]==0)
            {
                click(i+1,j);
                num++;
            }
        }
    }
    bool mark=true;
    rep(j,1,5)if (a[5][j]==0)mark=false;
    if (mark)ans=min(ans,num);
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        rep(i,1,5)rep(j,1,5)scanf("%1d",&aa[i][j]);
        ans=INF;
        rep(i,0,31)
        {
            init();
            solve(i);
        }
        if (ans>6)W(-1);
        else W(ans);
    }
    return 0;
}