UVA 11464 Even Parity（枚舉遞推）

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11464 - Even ParityTime limit: 3.000 seconds

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

題意：給出一個n*n的01矩陣（每個元素非0即1），選擇盡量少的0變成1，使得每個元素的上下左右的元素紙盒均為偶數。如果無解，輸出-1.

分析：最容易想到的方法是枚舉每個數字“變”還是“不變”，最後判斷整個矩陣是否滿足條件。這樣做最多需要枚舉2^225種情況，難以承受。

注意到n只有15，每一行只有不超過2^15=32768種情況，所以可以枚舉第一行的情況。接下來可以根據第一行計算出第二行，根據第二行計算出第三行……這樣，總時間複雜度降為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不能變成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個元素之和            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不能變成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;}

UVA 11464 Even Parity（枚舉遞推）