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11464 – Even Parity
We have a grid of size N x 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 x 4:
|
1 |
0 |
1 |
0 |
The parity of each cell would be |
1 |
3 |
1 |
2 |
|
1 |
1 |
1 |
1 |
2 |
3 |
3 |
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 first 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 Output for 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 |
Case 1: 0 |
题意：给出一个n*n的01矩阵（每一个元素非0即1），选择尽量少的0变成1，使得每一个元素的上下左右的元素纸盒均为偶数。假设无解，输出-1.
分析：最easy想到的方法是枚举每一个数字“变”还是“不变”，最后推断整个矩阵是否满足条件。这样做最多须要枚举2^225种情况。难以承受。
注意到n仅仅有15，每一行仅仅有不超过2^15=32768种情况，所以能够枚举第一行的情况。
接下来能够依据第一行计算出第二行，依据第二行计算出第三行……这样，总时间复杂度降为O(2^n * n^2)。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 20;
const int INF = 0x3fffffff;
int n, A[N][N], B[N][N];
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; //1不能变成0
}
for(int r = 1; r < n; r++) {
for(int c = 0; c < n; c++) {
int sum = 0; //元素B[r-1][c]的上、左、右3个元素之和
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(B[r][c] == 0 && A[r][c] == 1) return INF; //1不能变成0
}
}
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, cas = 0;
scanf("%d", &T);
while(T--) {
scanf("%d", &n);
for(int r = 0; r < n; r++)
for(int c = 0; c < n; c++)
scanf("%d", &A[r][c]);
int ans = INF;
for(int i = 0; i < (1<<n); i++)
ans = min(ans, check(i));
if(ans == INF) ans = -1;
printf("Case %d: %d\n", ++cas, ans);
}
return 0;
}
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