解題區/解題區 - Zerojudge

a243. 第四題：點燈遊戲

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

點燈遊戲，在二維地圖上，點擊一個燈泡後，亮會變暗、暗會變亮，還有周圍的燈泡也會改變狀態。滿足 $| x1 − x2 | + | y1 − y2 | = d$ 的位置的燈泡也會隨之變化，注意到是曼哈頓距離等於，而非小於等於。

給一個初始盤面，請問點燈結果有沒有解。

Sample Input

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1 1 1
1
2 2 1
1 1
1 1
3 2 1
1 0 1
0 1 0
3 3 1
1 0 1
0 1 0
1 0 1
4 4 2
1 1 0 1
0 0 0 1
1 0 1 1
1 0 0 0
5 5 1
1 1 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
5 5 2
0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
11 11 3
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
11 11 3
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
13 13 7
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0

Sample Output

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1
1
0
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0
0
1
1
0
1

Solution

轉換成聯立方程組，當盤面是 $N \times M$，則用 $N \times M$ 個變數和 $N \times M$ 等式進行解方程。燈泡變換只有 1 -> 0 和 0 -> 1，利用高斯消去法解聯立時需要使用 XOR 取代原本的加減乘除計算。變數 $x(i, j) = 1$ 表示要 $(i, j)$ 按下，反之 0 表示不按，最後再模擬一次看能不能翻回全暗。

關於列方程式，下面是一個 $(2, 2)$ 位置被點開的影響周圍情況

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0100
1110
0100
0000

相同地，周圍也可以影響到自己，那麼方程式就是 $x(1, 2) + x(2, 1) + x(2, 2) + x(2, 3) + x(3, 2) = L(2, 2)$。其中 + 號相當於 XOR 計算，而 L(i, j) 表示初始盤面狀態，當左式等於右式相當於 L(i, j) 被點暗。

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#include <bits/stdc++.h> 
using namespace std;
class XOR_GAUSSIAN { // XOR Gaussian Elimination
public:
    static const int MAXN = 700;
    char mtx[MAXN][MAXN+1], varX[MAXN];
    int n;
    void compute() {
    	int row = 0, col = 0, arb = 0;
    	int equ = n, var = n;
    	while (row < equ && col < var) {
    		int c = row;
    		for (int i = row; i < equ; i++)
    			if (mtx[i][col])
    				c = i;
    		for (int i = 0; i <= var; i++)
                swap(mtx[c][i], mtx[row][i]);
    		if (mtx[row][col] == 0) {
    			col++, arb++;
    			continue;
    		}    		
    		for (int i = 0; i < equ; i++) {
    			if (i == row || mtx[i][col] == 0) continue;
    			for (int j = var; j >= 0; j--)
    				mtx[i][j] ^= mtx[row][j];
    		}
    		row++, col++;
    	}
    	memset(varX, 0, sizeof(varX));
    	for (int i = 0, j; i < equ; i++) {
    		if (mtx[i][var] == 0)
    			continue;
    		for (j = 0; j < var && mtx[i][j] == 0; j++);
    		varX[j] = mtx[i][var];
    	}
    }
    void init(int n) {
    	this->n = n;
    	for (int i = 0; i < n; i++)
    		for (int j = 0; j <= n; j++)
    			mtx[i][j] = 0;
    }
} gauss;
int main() {
    int n, m, d;
    int g[32][32];
    int cases = 0;
    while (scanf("%d %d %d", &m, &n, &d) == 3 && n) {
        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                scanf("%d", &g[i][j]);
        int N = n*m;
        gauss.init(N);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int x = i*m + j, tx, ty;
                gauss.mtx[x][N] = g[i][j];
                gauss.mtx[x][x] = 1;
                for (int dx = -d; dx <= d; dx++) {
                    int v = d - abs(dx);
                    int dy[2] = {-v, v}, dn = v == 0 ? 1 : 2;
                    for (int k = 0; k < dn; k++) {
                        tx = i + dx, ty = j + dy[k];
                        if (tx < 0 || ty < 0 || tx >= n || ty >= m)
                            continue;
                        gauss.mtx[x][tx*m+ty] = 1;
                    }
                }
            }
        }
        gauss.compute();
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (gauss.varX[i*m+j] == 0)
                    continue;
                g[i][j] ^= 1;
                int tx, ty;
                for (int dx = -d; dx <= d; dx++) {
                    int v = d - abs(dx);
                    int dy[2] = {-v, v}, dn = v == 0 ? 1 : 2;
                    for (int k = 0; k < dn; k++) {
                        tx = i + dx, ty = j + dy[k];
                        if (tx < 0 || ty < 0 || tx >= n || ty >= m)
                            continue;
                        g[tx][ty] ^= 1;
                    }
                }
            }
        }
        
        int ok = 1;
        for (int i = 0; i < n; i++)  {
            for (int j = 0; j < m; j++) { 
                ok &= g[i][j] == 0;
            }
        }
        printf("%d\n", ok);
    }
    return 0;
}
/*
2 3 1
1 0 1 0 0 0
3 3 2
1 0 0
0 0 0
0 0 0
*/