UVA 11464 - Even Parity(枚举方法）
| D |
Even Parity Input: Standard Input Output: Standard Output |
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：给定n*n矩阵，可以把0变成1，求最少变幻次数使得每个位置的上下左右之和为偶数。
思路:n为15，第一行状态最多2^15种，然后由前一行可以推出后一行，如此一来，时间复杂度为O(2^n) * (n^2).
代码：
#include <stdio.h>
#include <string.h>
#define min(a,b) (a)<(b)?(a):(b)
#define INF 0x3f3f3f3f
const int d[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const int N = 15;
int t, n, map[N][N], save[N][N];
void init() {
scanf("%d", &n);
for (int i = 0; i < n; i ++)
for (int j = 0; j < n; j ++) {
scanf("%d", &map[i][j]);
}
}
int Sum(int i, int j) {
int sum = 0;
for (int k = 0; k < 4; k ++) {
int xx = i + d[k][0];
int yy = j + d[k][1];
if (xx >= 0 && xx < n && yy >= 0 && yy < n) {
sum += save[xx][yy];
}
}
return sum;
}
int cal(int state) {
int count = 0;
memset(save, 0, sizeof(save));
for (int i = 0; i < n; i ++)
for (int j = 0; j < n; j ++)
save[i][j] = map[i][j];
for (int i = n - 1; i >= 0; i --)
if (state >= (1<<i)) {
if (!save[0][n - 1 - i])
count ++;
save[0][n - 1 - i] = 1;
state -= (1<<i);
}
for (int i = 0; i < n - 1; i ++) {
for (int j = 0; j < n; j ++) {
if (Sum(i, j) % 2) {
if (save[i + 1][j])
return count = INF;
save[i + 1][j] = 1;
count ++;
}
}
}
for (int j = 0; j < n; j ++)
if (Sum(n - 1, j) % 2) {
count = INF;
break;
}
return count;
}
int judge(int state) {
for (int i = 0; i < n; i ++)
if (map[0][i] == 1 && (state&(1<<i) == 0))
return false;
return true;
}
void solve() {
int m = 1<<n, ans = INF;
for (int i = 0; i < m; i ++) {
if (judge(i)) {
ans = min(ans, cal(i));
}
}
if (ans == INF)
printf("-1\n");
else
printf("%d\n", ans);
}
int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
init();
printf("Case %d: ", ++cas);
solve();
}
return 0;
}