状压dp

题目链接 https://vjudge.net/contest/305270#problem/G
不太明算出来这个点的周围的1的个数后的操作，这一块 ，

b[i][j]=sum&1;
			if (a[i][j]==1 && b[i][j]==0) return inf;

#include 
using namespace std;
const int inf=999999999;
int a[20][20],b[20][20],n;
int check(int s);
int main()
{
	int t;
	scanf ("%d",&t);
	int cas=0;
	while (cas<t){
		scanf ("%d",&n);
		for (int i=0; i<n; i++)
		 for (int j=0; j<n; j++)
		  scanf ("%d",&a[i][j]);
		int ans=inf;
		for (int i=0; i<(1<<n); i++){
			ans=min(ans,check(i));
		}
		printf ("Case %d: ",++cas);
		if (ans==inf) printf ("-1\n");
		else printf ("%d\n",ans);
	}
	return 0;
}
int check(int s)
{
	memset(b,0,sizeof(b));
	int ans=0;
	for (int i=0; i<n; i++){
		if (s&(1<<i)) b[0][i]=1;
		else if (a[0][i]) return inf;
	}
	for (int i=1; i<n; i++){//row
		for (int j=0; j<n; j++){//col
			int sum=0;
			if(i>0) sum+=b[i-1][j];
			if(i<n-1) sum+= b[i+1][j];
			if (j>0) sum+=b[i-1][j-1];
			if (j<n-1) sum+=b[i-1][j+1];
			b[i][j]=sum&1;
			if (a[i][j]==1 && b[i][j]==0) return inf;
		}
	}
	for (int i=0; i<n; i++)
	    for (int j=0; j<n; j++){
	    	if (a[i][j]!=b[i][j]) ans++;
		}
	return ans;
}

还可以用dfs做

#include
#include
#include
using namespace std;
const int oo=0x3f3f3f3f;
int a[20][20],n,ans,tem[20][20];
int cal(int i,int j)
{
    int ret=0;
    if (i>1) ret+=tem[i-1][j];
    if (i<n) ret+=tem[i+1][j];
    if (j>1) ret+=tem[i][j-1];
    if (j<n) ret+=tem[i][j+1];
    return ret;
}
int solve()
{
    memcpy(tem,a,sizeof(a));
    int i,j,ret=0;
    for (i=2;i<=n;i++)
      for (j=1;j<=n;j++)
        if (cal(i-1,j)&1)
        {
            if (!tem[i][j])
            {
                tem[i][j]=1;
                ret++;
            }
            else return oo;
        }
    for (i=1;i<=n;i++)
      if (cal(n,i)&1) return oo;
    return ret;
}
void dfs(int p,int now)
{
    if (p==n+1)
    {
        ans=min(ans,now+solve());
        return;
    }
    dfs(p+1,now);
    if (!a[1][p])
    {
        a[1][p]=1;
        dfs(p+1,now+1);
        a[1][p]=0;
    }
}
int main()
{
    int T,K,i,j;
    scanf("%d",&T);
    for (K=1;K<=T;K++)
    {
        scanf("%d",&n);
        for (i=1;i<=n;i++)
          for (j=1;j<=n;j++)
            scanf("%d",&a[i][j]);
        ans=oo;
        dfs(1,0);
        printf("Case %d: ",K);
        if (ans<oo) printf("%d\n",ans);
        else printf("-1\n");
    }
}