UVA - 11464 Even Parity

11464 Even Parity
We have a grid of size N  N. Each cell of the grid initially contains a zero(0) or a one(1). The parity
of a cell is the number of 1s surrounding that cell. A cell is surrounded by at most 4 cells (top, bottom,
left, right).
Suppose we have a grid of size 4  4:
1 0 1 0 The parity of each cell would be 1 3 1 2
1 1 1 1 2 3 2 1
0 1 0 0 2 1 2 1
0 0 0 0 0 1 0 0
For this problem, you have to change some of the 0s to 1s so that the parity of every cell becomes
even. We are interested in the minimum number of transformations of 0 to 1 that is needed to achieve
the desired requirement.
Input
The rst line of input is an integer T (T < 30) that indicates the number of test cases. Each case starts
with a positive integer N (1 N 15). Each of the next N lines contain N integers (0/1) each. The
integers are separated by a single space character.
Output
For each case, output the case number followed by the minimum number of transformations required.
If it's impossible to achieve the desired result, then output `-1' instead.
Sample Input
3
3
0 0 0
0 0 0
0 0 0
3
0 0 0
1 0 0
0 0 0
3
1 1 1
1 1 1
0 0 0
Sample Output
Case 1: 0
Case 2: 3

Case 3: -1

pdf格式，效果较渣，见谅见谅。

刘汝佳大白书里的题，具体分析见大白书。

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=20;
const int inf=0x7fffffff;
int n,a[maxn][maxn],b[maxn][maxn];

int check(int s)
{
    memset(b,0,sizeof(b));
    for(int c=0;c<n;c++)
    {
        if(s&(1<<c)) b[0][c]=1;
        else if(a[0][c]==1) return inf;
    }
    for(int r=1;r<n;r++)
    for(int c=0;c<n;c++)
    {
        int sum=0;
        if(r>1) sum+=b[r-2][c];
        if(c>0) sum+=b[r-1][c-1];
        if(c<n-1) sum+=b[r-1][c+1];
        b[r][c]=sum%2;
        if(a[r][c]==1&&b[r][c]==0)return inf;
    }
    int cnt=0;
    for(int r=0;r<n;r++)
    for(int c=0;c<n;c++)
    if(a[r][c]!=b[r][c])
    cnt++;
    return cnt;
}
int main()
{
    int t,kase,r,c,s;
    scanf("%d",&t);
    for(kase=1;kase<=t;kase++)
    {
        scanf("%d",&n);
        for(r=0;r<n;r++)
        for(c=0;c<n;c++)
        scanf("%d",&a[r][c]);
        int ans=inf;
        for(s=0;s<(1<<n);s++)
        ans=min(ans,check(s));
        if(ans==inf)ans=-1;
        printf("Case %d: %d\n",kase,ans);
    }
    return 0;
}

题目不太难。哈哈，我又重新开始做acm了！梦想是什么？也许就是如果你一段时间不做，就会觉得恍然若失吧。再次提交，

看到ac，还是很有感触的。以后加油。还有就是我今天意外的联系到了2013年世界总决赛的世界亚军（郭晓旭），他是我高

中时候的学长。然后又意外的发现了好几个潮实或潮汕的acmer。呵呵，感觉还不错。