UVa11464 Even Parity

很有意思的题目，简单粗暴

根据大白上的思路来的，但是数组下标从1开始会好处理一点

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 20;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
typedef long long LL;
int n,a[maxn][maxn],b[maxn][maxn];
int solve(int s)
{
    memset(b,0,sizeof(b));
    for(int i = 1;i<=n;++i)
    {
        if(s & (1<<(i-1)))b[1][i] = 1;
        else if(a[1][i]==1)return INF;
    }
    for(int i = 1;i<n;++i)
        for(int j = 1;j<=n;++j)
        {
            if(a[i+1][j])
            {
                if((b[i][j-1]+b[i][j+1]+b[i-1][j]+1)%2)return INF;
                else {b[i+1][j] = 1;continue;}
            }
            b[i+1][j] = (b[i][j-1]+b[i][j+1]+b[i-1][j])%2?1:0;
        }

    int ans = 0;
    for(int i = 1;i<=n;++i)
        for(int j = 1;j<=n;++j)
            if(a[i][j]!=b[i][j])ans++;
//    printf("ans = %d\n",ans);
    return ans;
}
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    int T;scanf("%d",&T);
    for(int kase = 1;kase<=T;++kase)
    {
        scanf("%d",&n);
        for(int i = 1;i<=n;++i)
            for(int j = 1;j<=n;++j)
                scanf("%d",&a[i][j]);
        int ans = INF;
        for(int i = 0;i<(1<<n);++i)
            ans = min(ans,solve(i));
        if(ans>=INF)ans = -1;
        printf("Case %d: %d\n",kase,ans);
    }
    return 0;
}