POJ 1222 EXTENDED LIGHTS OUT(翻转+二维开关问题)

 POJ 1222 EXTENDED LIGHTS OUT

　　　　今天真是完美的一天,这是我在poj上的100A,留个纪念,马上就要期中考试了,可能后面几周刷题就没这么快了,不管怎样,为下一个200A奋斗,

　　　　这个题是大白上的牛翻转颜色的题(P153)的弱化版,典型的开关问题;

/*
* Created:     2016年04月05日 22时28分26秒 星期二
* Author:      Akrusher
*
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define in(n) scanf("%d",&(n))
#define in2(x1,x2) scanf("%d%d",&(x1),&(x2))
#define in3(x1,x2,x3) scanf("%d%d%d",&(x1),&(x2),&(x3))
#define inll(n) scanf("%I64d",&(n))
#define inll2(x1,x2) scanf("%I64d%I64d",&(x1),&(x2))
#define inlld(n) scanf("%lld",&(n))
#define inlld2(x1,x2) scanf("%lld%lld",&(x1),&(x2))
#define inf(n) scanf("%f",&(n))
#define inf2(x1,x2) scanf("%f%f",&(x1),&(x2))
#define inlf(n) scanf("%lf",&(n))
#define inlf2(x1,x2) scanf("%lf%lf",&(x1),&(x2))
#define inc(str) scanf("%c",&(str))
#define ins(str) scanf("%s",(str))
#define out(x) printf("%d
",(x))
#define out2(x1,x2) printf("%d %d
",(x1),(x2))
#define outf(x) printf("%f
",(x))
#define outlf(x) printf("%lf
",(x))
#define outlf2(x1,x2) printf("%lf %lf
",(x1),(x2));
#define outll(x) printf("%I64d
",(x))
#define outlld(x) printf("%lld
",(x))
#define outc(str) printf("%c
",(str))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mem(X,Y) memset(X,Y,sizeof(X));
typedef vector<int> vec;
typedef long long ll;
typedef pair<int,int> P;
const int dx[5]={-1,0,0,0,1},dy[5]={0,-1,0,1,0};
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;

const int M=5;
const int N=6;
int tile[5][6];
int opt[5][6];
int flip[5][6];
int k;
int get(int x,int y){
    int c=tile[x][y];
    rep(i,0,5){
        int nx=x+dx[i],ny=y+dy[i];
        if(nx>=0&&ny>=0&&nx<M&&ny<N){
            c+=flip[nx][ny];    
        }
    }
  return c%2;
 }
int calc(){
     rep(i,1,M){
         rep(j,0,N){
             if(get(i-1,j)) flip[i][j]=1;
             }
         }
     rep(j,0,N){
         if(get(M-1,j)){
             return -1;
         }
     }
     int res=0;
     rep(i,0,M) 
         rep(j,0,N)
             if(flip[i][j]) res++;
     return res;
}
void solve(){
    int res=-1;
    rep(i,0,1<<N){
        mem(flip,0);
        rep(j,0,N){
            flip[0][N-j-1]=(i>>j)&1;
        }
    int num=calc();
    if(num>=0&&(res<0||res>num)){ //此处可能为0
        res=num;
        memcpy(opt,flip,sizeof(flip));
    }
    }
    printf("PUZZLE #%d
",k);
    rep(i,0,M){
        rep(j,0,N){
            printf("%d ",opt[i][j]);
            }
        printf("
");
        }
}
int main()
{
    int t;
    in(t);
    k=0;
    while(t--){
        k++;
        rep(i,0,M)
            rep(j,0,N){
            in(tile[i][j]);
            }
        solve();
    }
    return 0;
}