POJ-1681 Painter's Problem 高斯消元
2013-05-30 22:24
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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 )。
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 )。
//STATUS:C++_AC_16MS_496KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream> #include <iomanip> #include <numeric> #include <cstring> #include <cassert> #include <cstdio> #include <string> #include <vector> #include <bitset> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <list> #include <set> #include <map> using namespace std; //using namespace __gnu_cxx; //define #define pii pair<int,int> #define mem(a,b) memset(a,b,sizeof(a)) #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define PI acos(-1.0) //typedef typedef long long LL; typedef unsigned long long ULL; //const const int N=300; const int INF=0x3f3f3f3f; const int MOD=100000,STA=8000010; const LL LNF=1LL<<60; const double EPS=1e-8; const double OO=1e15; const int dx[4]={-1,0,1,0}; const int dy[4]={0,1,0,-1}; const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; //Daily Use ... inline int sign(double x){return (x>EPS)-(x<-EPS);} template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} template<class T> inline T lcm(T a,T b,T d){return a/d*b;} template<class T> inline T Min(T a,T b){return a<b?a:b;} template<class T> inline T Max(T a,T b){return a>b?a:b;} template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} //End char ma ; int A ,B ,vis ,num ; int T,n; void getA(int n,int m) { int i,j,k,x,y; mem(A,0); for(i=0;i<n;i++){ for(j=0;j<m;j++){ A[i*m+j][i*m+j]=1; for(k=0;k<4;k++){ x=i+dx[k]; y=j+dy[k]; if(x>=0&&x<n && y>=0&&y<m){ A[i*m+j][x*m+y]=1; } } } } for(i=0;i<n;i++){ for(j=0;j<m;j++) A[i*m+j][n*m]=ma[i][j]; } } int gauss(int n) { int i,j,k,cnt,row,ok,ret,up,free; for(i=row=0;i<n;i++){ if(!A[row][i]){ for(j=row+1;j<n;j++){ if(A[j][i]){ for(k=i;k<=n;k++)swap(A[row][k],A[j][k]); break; } } } if(A[row][i]!=1)continue; for(j=0;j<n;j++){ if(j!=row && A[j][i]){ for(k=i;k<=n;k++) A[j][k]^=A[row][k]; } } row++; } for(i=n-1;i>=row;i--) if(A[i] )return -1; if(row==n){ ret=0; for(i=0;i<n;i++)if(A[i] )ret++; return ret; } mem(vis,0); for(i=k=j=0;i<row;i++){ while(!A[i][j] && j<n){ vis[j]=1; num[k++]=j++; } } ret=INF;free=n-row; up=1<<free; for(k=0;k<up;k++){ for(i=0;i<free;i++)B[num[i]]=(k&(1<<i))?1:0; for(i=n-1;i>=0;i--){ if(!vis[i])continue; B[i]=0; for(j=row;j<n;j++)B[i]^=B[j]*A[i][j]; B[i]^=A[i] ; } for(i=cnt=0;i<n;i++)if(B[i])cnt++; ret=Min(ret,cnt); } return ret; } int main() { // freopen("in.txt","r",stdin); int i,j,ans; scanf("%d",&T); while(T--) { scanf("%d",&n); for(i=0;i<n;i++){ scanf("%s",ma[i]); for(j=0;j<n;j++)ma[i][j]=(ma[i][j]=='y'?0:1); } getA(n,n); ans=gauss(n*n); if(ans>=0)printf("%d\n",ans); else printf("inf\n"); } return 0; }
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