BZOJ3503: [Cqoi2014]和谐矩阵

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=3503

这种写法很慢。。。

每个点建立一个xor方程，分别对它上下左右还有自己的系数都为1其他都为0，然后自由元就都设为1。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#define rep(i,l,r) for (int i=l;i<=r;i++)
#define down(i,l,r) for (int i=l;i>=r;i--)
#define clr(x,y) memset(x,y,sizeof(x))
#define maxn 2000
using namespace std;
int n,m,a[maxn][maxn],ans[maxn];
char ch;
int dx[5]={0,0,1,0,-1},dy[5]={0,1,0,-1,0};
int read(){
    int x=0,f=1; char ch=getchar();
    while (!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
    while (isdigit(ch)) {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}
bool jud(int x,int y){
    if (x<1||x>n||y<1||y>m) return 0;
    return 1;
}
int p(int x,int y){
    return (x-1)*m+y;
}
void gs(int n){
    rep(i,1,n){
        int j=i;
        while (!a[j][i]&&j<=n) j++; 
        if (j>n) continue;
        if (i!=j) rep(k,1,n+1) swap(a[i][k],a[j][k]); 
        rep(j,i+1,n) if (j!=i&&a[j][i])
            rep(k,1,n+1) a[j][k]^=a[i][k];  
    }
    down(i,n,1){
        if (!a[i][i]) {ans[i]=1; continue;}
        int ret=a[i][n+1];
        down(j,n,i+1) if (a[i][j]) ret^=ans[j];
        ans[i]=ret;
    }
}
int main(){
  //    freopen("in.txt","r",stdin);
    n=read(); m=read();
    rep(i,1,n) rep(j,1,m) rep(k,0,4){
        int x=i+dx[k],y=j+dy[k];
        if (jud(x,y)) a[p(i,j)][p(x,y)]=1;
    }
    gs(n*m);
    int cnt=0;
    rep(i,1,n) {
        rep(j,1,m) printf("%d ",ans[++cnt]);
        puts("");
    }
    return 0;
}