CF1157G. Inverse of Rows and Columns

Solution

首先枚举第一行是否变换，再枚举第一行的$01$状态，即可确定列变换。
然后对于之后的行变换，从前往后贪心地让$1$出现得尽可能晚即可。

Code

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se secondusing namespace std;template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=1e9+7;
const int MAXN=205;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{int f=1,x=0; char c=getchar();while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }return x*f;
}
int a[MAXN][MAXN],X[MAXN],Y[MAXN],Flag[MAXN];
signed main()
{int n=read(),m=read();for (int i=1;i<=n;i++)for (int j=1;j<=m;j++) a[i][j]=read();for (int i=0;i<=m;i++){int flag=1;for (int j=1;j<=i;j++) Y[j]=a[1][j]^X[1];for (int j=i+1;j<=m;j++) Y[j]=a[1][j]^X[1]^1;for (int j=1;j<=n;j++) Flag[j]=X[j]=0;for (int j=2,lst=(i!=m);j<=n;j++){for (int k=1;k<=m;k++)if (!lst){if (Flag[j]&&(a[j][k]^X[j]^Y[k])) lst=1;if (!Flag[j]&&k>1&&(a[j][k]^X[j]^Y[k])) lst=1;if (!Flag[j]&&k==1&&(a[j][k]^X[j]^Y[k])) X[j]=Flag[j]=1;}else{if (Flag[j]&&!(a[j][k]^X[j]^Y[k])) flag=0;if (!Flag[j]&&(a[j][k]^X[j]^Y[k])) X[j]=0,Flag[j]=1;if (!Flag[j]&&!(a[j][k]^X[j]^Y[k])) X[j]=1,Flag[j]=1;}}if (flag){puts("YES");for (int j=1;j<=n;j++) putchar(X[j]+'0');puts("");for (int j=1;j<=m;j++) putchar(Y[j]+'0');puts("");return 0;} }X[1]=1;for (int i=0;i<=m;i++){int flag=1;for (int j=1;j<=i;j++) Y[j]=a[1][j]^X[1];for (int j=i+1;j<=m;j++) Y[j]=a[1][j]^X[1]^1;for (int j=1;j<=n;j++) Flag[j]=X[j]=0;for (int j=2,lst=(i!=m);j<=n;j++){for (int k=1;k<=m;k++)if (!lst){if (Flag[j]&&(a[j][k]^X[j]^Y[k])) lst=1;if (!Flag[j]&&k>1&&(a[j][k]^X[j]^Y[k])) lst=1;if (!Flag[j]&&k==1&&(a[j][k]^X[j]^Y[k])) X[j]=Flag[j]=1;}else{if (Flag[j]&&!(a[j][k]^X[j]^Y[k])) flag=0;if (!Flag[j]&&(a[j][k]^X[j]^Y[k])) X[j]=0,Flag[j]=1;if (!Flag[j]&&!(a[j][k]^X[j]^Y[k])) X[j]=1,Flag[j]=1;}}if (flag){puts("YES");for (int j=1;j<=n;j++) putchar(X[j]+'0');puts("");for (int j=1;j<=m;j++) putchar(Y[j]+'0');puts("");return 0;} }puts("NO");return 0;
}