Codeforces #663 div.2 D

题目

A binary matrix is called good if every even length square sub-matrix has an odd number of ones.

Given a binary matrix a consisting of n rows and m columns, determine the minimum number of cells you need to change to make it good, or report that there is no way to make it good at all.

All the terms above have their usual meanings — refer to the Notes section for their formal definitions.

Input
The first line of input contains two integers n and m (1≤n≤m≤106 and n⋅m≤106) — the number of rows and columns in a, respectively.

The following n lines each contain m characters, each of which is one of 0 and 1. If the j-th character on the i-th line is 1, then ai,j=1. Similarly, if the j-th character on the i-th line is 0, then ai,j=0.

Output
Output the minimum number of cells you need to change to make a good, or output −1 if it's not possible at all.

Examples
inputCopy
3 3
101
001
110
outputCopy
2
inputCopy
7 15
000100001010010
100111010110001
101101111100100
010000111111010
111010010100001
000011001111101
111111011010011
outputCopy
-1

思路

显然当n和m有一个为1或两者都大于4那么都不成立（前者为0），那么数据范围就被压缩到了2到3之间，因为很小，所以我们可以预处理出来合法的相邻两列的状态，然后状态dp去枚举，做法类似于炮兵阵地。

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x7f7f7f7f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
typedef long long ll;
typedef pair<int ,int> PII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }

const int maxn=1000005;
int n,m;
int ans=inf,dp[maxn][8],a[4][maxn];
bool st[8][8]={false};
const int mod=1e9+7;

int calc (int x) {
    int ans=0;
    while (x) {
        if (x&1) ans++;
        x>>=1;
    }
    return ans;
}

int main () {
//    freopen ("data.in","r",stdin);
   cin>>n>>m;
   if (n>=4&&m>=4) {
       cout<<"-1"<<endl;
       return 0;
   }
   if (n==1||m==1) {
       cout<<0<<endl;
       return 0;
   }
   rep (i,1,n)
    rep (j,1,m) scanf ("%1d",&a[i][j]);
   rev (i,0,8) 
    rev (j,0,8) {
        int t[3];
        t[0]=t[1]=t[2]=0;
        rev (p,0,n) {
            if ((1<<p)&i) t[p]++;
            if ((1<<p)&j) t[p]++;
        }
        bool flag=1;
        rev (p,0,n-1) {
            if (!((t[p]+t[p+1])&1)) flag=0;
        }
        if (flag) st[i][j]=1;
    } 
    MT (dp,0x7f7f7f7f);
    rev (i,0,8) dp[0][i]=0;
    rep (i,1,m) {
        int now=a[1][i]+a[2][i]*2+a[3][i]*4;
        rev (j,0,8) {
            rev (p,0,8) {
                if (!st[j][p]) continue;
                dp[i][j]=min (dp[i][j],dp[i-1][p]+calc (now^j));
            }
        }
    }
    rev (i,0,8) ans=min (ans,dp[m][i]);
    printf ("%d\n",ans);
   
//    fclose (stdin);
   return 0;
}