POJ 1681 Painter's Problem 高斯消元 枚举自由变量

题目链接：http://poj.org/problem?id=1681

和POJ 1753 一样

代码：

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#define sf scanf#define pf printfusing namespace std;const int maxn = 400,INF = 0x3f3f3f3f;typedef int Matrix[maxn][maxn];const int step[][2] = { {0,0} , {-1,0} , {1,0} , {0,1} , {0,-1} };Matrix A;int ANS[maxn],n;void init(){    memset(A,0,sizeof A);    int i,j,k,c_pos,n_pos,cx,cy,nx,ny;    for(i = 0;i < n;++i){        for(j = 0;j < n;++j){            c_pos = i * n + j;            for(k = 0;k < 5;++k){                nx = i + step[k][0],ny = j + step[k][1];                if(nx >= 0 && nx < n && ny >= 0 && ny < n){                    A[nx * n + ny][c_pos] = 1;                }            }        }    }}int Gauss(Matrix A,int n){    int i,j,k,p;    for(i = j = 0;j < n && i < n;++j){        k = i;        for(p = i + 1;p < n;++p) if(A[p][j] > A[k][j]) k = p;        if(A[k][j]){            for(p = 0;p <= n;++p) swap(A[i][p],A[k][p]);            for(k = i + 1;k < n;++k){                if(A[k][j]){                    for(p = j;p <= n;++p) A[k][p] = (A[k][p] - A[i][p] + 2) % 2;                }            }            i++;        }    }    for(k = i;k < n;++k) if(A[k][n]) return INF;    int cnt = n - i,minAns = INF,tmpAns;    for(int st = 0;st < (1 << cnt);++st){        tmpAns = 0;        for(j = i;j < n;++j) tmpAns += (ANS[j] = ((st >> (n - j - 1)) & 1));        for(j = i - 1;j >= 0;--j){            int tmp = 0;            for(p = j + 1;p < n;++p) tmp = (tmp + A[j][p] * ANS[p]) % 2;            ANS[j] = ( (A[j][n] - tmp) % 2 + 2 ) % 2;            tmpAns += ANS[j];        }        minAns = min(tmpAns,minAns);    }    return minAns;}char tmpc;int main(){    int T;sf("%d",&T);    while(T--){        sf("%d",&n);        init();getchar();        for(int i = 0;i < n * n;++i){            while((tmpc = getchar()) == '\n');            if(tmpc == 'y') A[i][n * n] = 0;            else A[i][n * n] = 1;        }        int ans = Gauss(A,n * n);        if(ans == INF) pf("inf\n");        else pf("%d\n",ans);    }}