POJ1681 Painter's Problem 高斯消元

　　异或高斯消元。如果是唯一解，则直接拿解与初始状态比较。如果有多解，则枚举自由变元的的取值情况，最坏复杂度O( 2^N )。
  1 //STATUS:C++_AC_16MS_496KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 typedef long long LL;
 33 typedef unsigned long long ULL;
 34 //const
 35 const int N=300;
 36 const int INF=0x3f3f3f3f;
 37 const int MOD=100000,STA=8000010;
 38 const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 char ma[N][N];
 58 int A[N][N],B[N],vis[N],num[N];
 59 int T,n;
 60 
 61 void getA(int n,int m)
 62 {
 63     int i,j,k,x,y;
 64     mem(A,0);
 65     for(i=0;i<n;i++){
 66         for(j=0;j<m;j++){
 67             A[i*m+j][i*m+j]=1;
 68             for(k=0;k<4;k++){
 69                 x=i+dx[k];
 70                 y=j+dy[k];
 71                 if(x>=0&&x<n && y>=0&&y<m){
 72                     A[i*m+j][x*m+y]=1;
 73                 }
 74             }
 75         }
 76     }
 77     for(i=0;i<n;i++){
 78         for(j=0;j<m;j++)
 79             A[i*m+j][n*m]=ma[i][j];
 80     }
 81 }
 82 
 83 int gauss(int n)
 84 {
 85     int i,j,k,cnt,row,ok,ret,up,free;
 86     for(i=row=0;i<n;i++){
 87         if(!A[row][i]){
 88             for(j=row+1;j<n;j++){
 89                 if(A[j][i]){
 90                     for(k=i;k<=n;k++)swap(A[row][k],A[j][k]);
 91                     break;
 92                 }
 93             }
 94         }
 95         if(A[row][i]!=1)continue;
 96         for(j=0;j<n;j++){
 97             if(j!=row && A[j][i]){
 98                 for(k=i;k<=n;k++)
 99                     A[j][k]^=A[row][k];
100             }
101         }
102         row++;
103     }
104     for(i=n-1;i>=row;i--)
105         if(A[i][n])return -1;
106     if(row==n){
107         ret=0;
108         for(i=0;i<n;i++)if(A[i][n])ret++;
109         return ret;
110     }
111     mem(vis,0);
112     for(i=k=j=0;i<row;i++){
113         while(!A[i][j] && j<n){
114             vis[j]=1;
115             num[k++]=j++;
116         }
117     }
118     ret=INF;free=n-row;
119     up=1<<free;
120     for(k=0;k<up;k++){
121         for(i=0;i<free;i++)B[num[i]]=(k&(1<<i))?1:0;
122         for(i=n-1;i>=0;i--){
123             if(!vis[i])continue;
124             B[i]=0;
125             for(j=row;j<n;j++)B[i]^=B[j]*A[i][j];
126             B[i]^=A[i][n];
127         }
128         for(i=cnt=0;i<n;i++)if(B[i])cnt++;
129         ret=Min(ret,cnt);
130     }
131     return ret;
132 }
133 
134 int main()
135 {
136  //   freopen("in.txt","r",stdin);
137     int i,j,ans;
138     scanf("%d",&T);
139     while(T--)
140     {
141         scanf("%d",&n);
142         for(i=0;i<n;i++){
143             scanf("%s",ma[i]);
144             for(j=0;j<n;j++)ma[i][j]=(ma[i][j]=='y'?0:1);
145         }
146         getA(n,n);
147 
148         ans=gauss(n*n);
149         if(ans>=0)printf("%d\n",ans);
150         else printf("inf\n");
151     }
152     return 0;
153 }