hdu 3364 Lanterns 高斯消元，模板题

#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <queue>#include <stack>#include <iostream>#include <algorithm>using namespace std;const int maxn=55;struct matrix{    int f[maxn][maxn];}e,g;int find(matrix a,int m,int n)//高斯消元{    int i=1,j=1,k,r,u;    while(i<=m&&j<=n)//处理第i个方程，第j个变量    {        r=i;        for(k=i;k<=m;k++)        if(a.f[k][j]){r=k;break;}        if(a.f[r][j])        {            if(r!=i)for(k=0;k<=n+1;k++)swap(a.f[r][k],a.f[i][k]);            for(u=i+1;u<=m;u++)if(a.f[u][j])                for(k=i;k<=n+1;k++)a.f[u][k]^=a.f[i][k];            i++;        }        j++;    }    for(u=i;u<=m;u++)//判断无解        if(a.f[u][n+1])return -1;    return i-1;}int main(){    int T,tt=0;    cin>>T;    while(T--)    {        int i,j,k,n,m,a,b,q,r;        cin>>n>>m;        memset(e.f,0,sizeof(e.f));        for(i=1;i<=m;i++)        {            cin>>k;            for(j=0;j<k;j++)            {                cin>>a;                e.f[a][i]=1;            }        }        cout<<"Case "<<++tt<<":"<<endl;        cin>>q;        while(q--)        {            for(i=1;i<=n;i++)            {                cin>>a;                e.f[i][m+1]=a;            }            r=find(e,n,m);//求出有界遍历个数r            if(r==-1)cout<<0<<endl;            else cout<<(1LL<<(m-r))<<endl;//注意范围，超int        }    }    return  0;}/*    设m个开关状态为xi(1<=i<=m,0<=xi<=1)，可列m元一次方程组。用高斯消元法求得自由变量个数k。答案就是2^k*/