uva 1560 - Extended Lights Out(枚举 | 高斯消元)

题目链接：uva 1560 - Extended Lights Out

题目大意：给定一个5∗6的矩阵，每一个位置上有一个灯和开关，初始矩阵表示灯的亮暗情况，假设按了这个位置的开关，将会导致周围包含自己位置的灯状态变换。求一个按开关位置，保证全部灯都灭掉。

解题思路：

1. 枚举，枚举第一行的状态，然后递推出后面四行的状态。
2.  高斯消元，对于每一个位置对定变量，这样列出30个方程求解。

C++ 枚举
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 10;
const int R = 5;
const int C = 6;

int arr[maxn][maxn], v[maxn][maxn], p[maxn][maxn];

bool judge (int s) {
    memset(p, 0, sizeof(p));
    memset(v, 0, sizeof(v));

    for (int i = 0; i < C; i++) {
        if (s&(1<<i)) {
            p[1][i+1] = 1;

            for (int j = 0; j <= 2; j++)
                v[1][i+j] ^= 1;
            v[2][i+1] ^= 1;
        }
    }

    for (int i = 2; i <= R; i++) {
        for (int j = 1; j <= C; j++)
            if (v[i-1][j]^arr[i-1][j]) {
                p[i][j] = 1;

                for (int k = -1; k <= 1; k++)
                    v[i][j+k] ^= 1;
                v[i+1][j] ^= 1;
            }
    }

    for (int i = 1; i <= C; i++)
        if (v[R][i]^arr[R][i])
            return false;

    for (int i = 1; i <= R; i++) {
        for (int j = 1; j < C; j++)
            printf("%d ", p[i][j]);
        printf("%d\n", p[i][C]);
    }
    return true;
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {

        for (int i = 1; i <= R; i++)
            for (int j = 1; j <= C; j++)
                scanf("%d", &arr[i][j]);

        printf("PUZZLE #%d\n", kcas);
        for (int s = 0; s < (1<<C); s++)
            if (judge(s))
                break;
    }
    return 0;
}

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int dir[5][2] = {{0, 0}, {1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const int maxn = 30;
const int R = 5;
const int C = 6;
typedef int Mat[maxn+5][maxn+5];

Mat A;
int v[R+5][C+5];

void init () {
    memset(A, 0, sizeof(A));
    memset(v, 0, sizeof(v));
    for (int i = 0; i < R; i++) {
        for (int j = 0; j < C; j++) {
            int x = i * C + j;
            scanf("%d", &A[x][maxn]);
            for (int k = 0; k < 5; k++) {
                int p = i + dir[k][0];
                int q = j + dir[k][1];
                if (p < 0 || p >= R || q < 0 || q >= C)
                    continue;
                A[x][p*C+q] = 1;
            }
        }
    }
}

void gauss_elimination (Mat a, int n) {

    for (int i = 0; i < n; i++) {

        int r = i;
        while (A[r][i] == 0)
            r++;

        if (r != i) {
            for (int j = 0; j <= n; j++)
                swap(A[i][j], A[r][j]);
        }

        for (int j = i + 1; j < n; j++) {
            if (A[j][i]) {
                for (int k = 0; k <= n; k++)
                    A[j][k] ^= A[i][k];
            }
        }
    }

    for (int i = n - 1; i >= 0; i--) {
        for (int j = i + 1; j < n; j++)
            A[i][n] ^= (A[j][n] * A[i][j]);
        if (A[i][n])
            v[i/C][i%C] = 1;
    }
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();
        gauss_elimination(A, maxn);
        printf("PUZZLE #%d\n", kcas);
        for (int i = 0; i < R; i++) {
            printf("%d", v[i][0]);
            for (int j = 1; j < C; j++)
                printf(" %d", v[i][j]);
            printf("\n");
        }
    }
    return 0;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。