Poj1222EXTENDED LIGHTS OUT高斯消元

题意：1表示开着，0表示关着。当对某个等打开或关闭时，周围四个灯的状态也会改变，让输出一种开关灯的方式。

搞法1：高斯消元，不过感觉数据应该有问题，我解方程是按一定会存在唯一一组解的情况写的。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL;
using namespace std;
int dx[4] = { -1, 0, 1, 0 };
int dy[4] = { 0, -1, 0, 1 };
int a[100][100];
void gao(int pos, int x, int y)
{
	a[pos][x * 6 + y] = 1;
	for (int i = 0; i < 4; i++){
		int xx = x + dx[i]; int yy = y + dy[i];
		if (xx >= 0 && xx < 5 && yy >= 0 && yy < 6){
			a[pos][xx * 6 + yy] = 1;
		}
	}
}



void Gauss(int equ, int var)
{
	int k; int col;
	for (k = 0, col = 0; k < equ&&col < var; k++, col++){
		int kk = k;
		for (int i = k + 1; i < equ; i++){
			if (a[i][col]>a[kk][col]) kk = i;
		}
		if (kk != k){
			for (int j = k; j <= var; j++)
				swap(a[k][j], a[kk][j]);
		}
		if (a[k][col] == 0){
			k--; continue;
		}
		for (int i = 0; i < equ; i++){
            if(i==k) continue;
			if (a[i][col]){
				for (int j = col; j <= var; j++){
					a[i][j] ^= a[k][j];
				}
			}
		}
	}
	for (int i = 0; i<30; i++){
		printf("%d ", a[i][30]);
		if ((i + 1) % 6 == 0) printf("
");
	}
}
int m[30][30];
int main()
{
	int T;
	cin >> T; int Icase = 0;
	while (T--){
		memset(a, 0, sizeof(a));
		printf("PUZZLE #%d
", ++Icase);
		for (int i = 0; i < 5; i++){
			for (int j = 0; j < 6; j++){
				scanf("%d", &m[i][j]);
				a[i * 6 + j][30] = m[i][j];
			}
		}
		for (int i = 0; i < 5; i++){
			for (int j = 0; j < 6; j++){
				gao(i * 6 + j, i, j);
			}
		}
		Gauss(30, 30);
	}
	return 0;
}

　　

搞法2：枚举第一行状态，也就（1<<6)种，下面的开关状态也就确定了。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <set>
#include <bitset>
#include <queue>
#include <stack>
#include <string>
#include <iostream>
#include <cmath>
#include <climits>
typedef long long LL;
using namespace std;
int dx[4] = { 0, 1, 0, -1 };
int dy[4] = { 1, 0, -1, 0 };
int m[10][10];
int m1[10][10];
int ans[10];
void nong(int x, int y)
{
	m1[x][y] ^= 1;
	for (int i = 0; i < 4; i++){
		int xx = x + dx[i]; int yy = y + dy[i];
		if (xx >= 0 && xx < 5 && yy >= 0 && yy < 6) m1[xx][yy] ^= 1;
	}
}

int gao(int x)
{
	memset(ans, 0, sizeof(ans));
	ans[0] = x;
	for (int i = 0; i < 5;i++)
	for (int j = 0; j < 6; j++) m1[i][j] = m[i][j];
	for (int i = 0; i < 6; i++){
		if (x&(1 << i)){
			nong(0, i);
		}
	}
	for (int i = 1; i < 5; i++){
		for (int j = 0; j < 6; j++){
			if (m1[i - 1][j]){
				ans[i] |= (1 << j);
				nong(i, j);
			}
		}
	}
	int flag = 1;
	for (int i = 0; i < 5&&flag; i++){
		for (int j = 0; j < 6&&flag; j++) if (m1[i][j]) flag = 0;
	}
	return flag;
}

int main()
{
	int T;
	cin >> T;
	int Icase = 0;
	while (T--){
		printf("PUZZLE #%d
", ++Icase);
		for (int i = 0; i < 5;i++)
		for (int j = 0; j < 6; j++) scanf("%d", &m[i][j]);
		for (int i = 0; i < (1 << 6); i++){
			int k = gao(i);
			if (k){
				for (int j = 0; j < 5; j++){
					for (int k = 0; k < 6; k++){
						if (ans[j] & (1 << k)) printf("1 ");
						else printf("0 ");
					}
					cout << endl;
				}
				break;
			}
		}
	}
	return 0;
}