UVA 11464 - Even Parity(枚举方法)

Description

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
Case 2: 3
Case 3: -1

                                                                                

题意：

给定n*n(1≤n≤15) 的矩阵，只可以从0变为1，求出最少的变换字数使得每一个数的上下左右和为偶数。

思路：

因为n 最大为15 ，所以第一行最多有2 ^15种情况，枚举第一行的情况，就可以由此计算后面的。利用位运算来枚举第一行的所有情况。

CODE;

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <string>#include <cstring>#include <queue>#include <stack>#include <vector>#include <set>#include <map>const int inf=0xfffffff;typedef long long ll;using namespace std;int a[20][20], b[20][20];int n;int cal(int s){    memset(b, 0, sizeof(b));    for(int i = 0; i < n; i ++){        if(s & (1<<i)) b[0][i] = 1;        else if(a[0][i] == 1) return inf;    }    for(int i = 1; i < n; i++){//算b[i-1][j]的上左右，从而得到b[i][j]应为多少。        for(int j = 0; j < n; j++){            int sum = 0;            if(i > 1) sum += b[i - 2][j];            if(j > 0 ) sum += b[i - 1][j - 1];            if(j < n-1) sum += b[i - 1][j + 1];            b[i][j] = sum % 2;            if(a[i][j] == 1 && b[i][j]== 0) return inf;        }    }    int cnt = 0;    for(int i = 0; i < n; i++){        for(int j = 0; j < n; j++)            if(a[i][j] != b[i][j]) cnt ++;    }    return cnt;}int main(){    //freopen("in", "r", stdin);    int T;    scanf("%d", &T);    for(int ca = 1; ca <= T; ca++){        scanf("%d", &n);        for(int i = 0; i < n; i++){            for(int j = 0; j < n; j++)                scanf("%d", &a[i][j]);        }        int ans = inf;        for(int i = 0; i < (1<<n); i++){            ans = min(ans, cal(i));        }        if(ans == inf) printf("Case %d: -1\n", ca);        else printf("Case %d: %d\n", ca, ans);    }    return 0;}