[SDOI 2010] Alien millipedes

Topic description:

QAQ.

Topic Analysis:

The observation coefficients are all 0 or 1.
This is called Gaussian elimination to solve the XOR equation.
Normal solution: a solution, the maximum number of rows visited is the answer, and then M[I][n+1] is the solution
No solution: that is, a row 1...n is all 0 but n+1 has a number (that is, 0x+0y+0z... = a non-zero number, which is obviously impossible), there is no such situation in this problem.
Countless solutions: that is, a row is all 0 (that is, 0x+0y+ 0z... = 0, obviously xyz can be taken arbitrarily)
The normal high consumption is O(N^3), which is theoretically impossible to passactually passed...
Considering that each elimination is the overall XOR of two lines, we can directly do a bitset optimization
O(N^2*m/64)

Topic link:

BZOJ 1923
Luogu 2447

AC code:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <bitset>
const int maxm=2100;
std::bitset<maxm> a[maxm];
int n,m;
inline int Gauss()
{
    int now=0,ans=0;
    for(int i=1;i<=n;i++)
    {
        int j=now+1;
        while(j<=m&&!a[j][i]) j++;
        if(j>m) return -1;
        ans=std::max(ans,j);
        now++;
        std::swap(a[now],a[j]);
        for(int j=1;j<= m;j++)
         if(j!=now&&a[j][i]) a[j]^=a[now];
    }
    return ans;
}
char s[maxm];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%s",s+1);
        for(int j=1;j<=n;j++)
         a[i][j]=s[j]-'0';
        scanf("%s",s+1);
        a[i][n+1]=s[1]-'0';
    }
    int d=Gauss();
    if(~d)
    {
        printf("%d\n",d);
        for(int i=1;i<=n;i++)
         if(!a[i][n+1]) puts("Earth");
         else puts("?y7M#");
    } 
    else puts("Cannot Determine");
    return 0;
}