POJ-1681 Painter's Problem 高斯消元

　　题目链接：http://poj.org/problem?id=1681

　　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 }