UVa 11464 - Even Parity
传送门
题意
给出一个原始矩阵,求最少数目变动,将0变为1,使得每个元素上下左右之和是偶数。
思路
用类似状态压缩的方法表示出第一行的状态,然后依次往下递推。
代码
#include <bits/stdc++.h>
#define LL long long
#define lowbit(x) ((x) & (-x))
const int MAXN = 15 + 5;
const int INF = 0x3f3f3f3f;
using namespace std;
int mp[MAXN][MAXN], n, tmp[MAXN][MAXN];
int Solve(int k)
{
memset(tmp, 0, sizeof tmp);
int i, j;
for (i = 0; i < n; i++)
if ((1 << i) & k)
tmp[0][i] = 1;
else if (mp[0][i] == 1)
return INF;
for (i = 1; i < n; i++)
{
for (j = 0; j < n; j++)
{
int cur = 0;
if (i > 1)
cur += tmp[i - 2][j];
if (j > 0)
cur += tmp[i - 1][j - 1];
if (j < n - 1)
cur += tmp[i - 1][j + 1];
tmp[i][j] = cur % 2;
if (tmp[i][j] == 0 && mp[i][j] == 1)
return INF;
}
}
int cnt = 0;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
if (tmp[i][j] != mp[i][j])
cnt++;
return cnt;
}
int main()
{
//freopen("input.txt", "r", stdin);
int T, i, j, cases = 0;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
int ans = INF;
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
scanf("%d", ∓[i][j]);
for (i = 0; i < (1 << n); i++)
ans = min(ans, Solve(i));
if (ans == INF) ans = -1;
printf("Case %d: %d\n", ++cases, ans);
}
return 0;
}
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