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\n",(x))
#define out2(x1,x2) printf("%d %d\n",(x1),(x2))
#define outf(x) printf("%f\n",(x))
#define outlf(x) printf("%lf\n",(x))
#define outlf2(x1,x2) printf("%lf %lf\n",(x1),(x2));
#define outll(x) printf("%I64d\n",(x))
#define outlld(x) printf("%lld\n",(x))
#define outc(str) printf("%c\n",(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[]={-,,,,},dy[]={,-,,,};
const int INF=0x3f3f3f3f;
const ll mod=1e9+;
ll powmod(ll a,ll b) {ll res=;a%=mod;for(;b;b>>=){if(b&)res=res*a%mod;a=a*a%mod;}return res;}
const bool AC=true;
 
const int M=;
const int N=;
int tile[][];
int opt[][];
int flip[][];
int k;
int get(int x,int y){
    int c=tile[x][y];
    rep(i,,){
        int nx=x+dx[i],ny=y+dy[i];
        if(nx>=&&ny>=&&nx<M&&ny<N){
            c+=flip[nx][ny];
        }
    }
  return c%;
 }
int calc(){
     rep(i,,M){
         rep(j,,N){
             if(get(i-,j)) flip[i][j]=;
             }
         }
     rep(j,,N){
         if(get(M-,j)){
             return -;
         }
     }
     int res=;
     rep(i,,M)
         rep(j,,N)
             if(flip[i][j]) res++;
     return res;
}
void solve(){
    int res=-;
    rep(i,,<<N){
        mem(flip,);
        rep(j,,N){
            flip[][N-j-]=(i>>j)&;
        }
    int num=calc();
    if(num>=&&(res<||res>num)){ //此处可能为0
        res=num;
        memcpy(opt,flip,sizeof(flip));
    }
    }
    printf("PUZZLE #%d\n",k);
    rep(i,,M){
        rep(j,,N){
            printf("%d ",opt[i][j]);
            }
        printf("\n");
        }
}
int main()
{
    int t;
    in(t);
    k=;
    while(t--){
        k++;
        rep(i,,M)
            rep(j,,N){
            in(tile[i][j]);
            }
        solve();
    }
    return ;
}