这题我们可以参考开关那题,只不过这里是求最少的操作次数,那么我们需要对变元进行枚举,算出所有的情况下,最少需要改变的次数。
代码如下:
#include <cstdlib> #include <cstring> #include <cstdio> #include <algorithm> #define TO(x, y) (x-1)*4+y using namespace std; char G[6][6]; int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1}; inline bool judge(int x, int y) { if (x < 1 || x > 4 || y < 1 || y > 4) { return false; } return true; } void swap(int &a, int &b) { int t = a; a = b; b = t; } struct Matrix { int a[20][20]; void init() { int xx, yy, k, pos; memset(a, 0, sizeof (a)); for (int i = 1; i <= 4; ++i) { for (int j = 1; j <= 4; ++j) { k = TO(i, j); a[k][k] = 1; for (int d = 0; d < 4; ++d) { xx = i + dir[d][0], yy = j + dir[d][1]; if (judge(xx, yy)) { pos = TO(xx, yy); a[pos][k] = 1; } } } } } void rswap(int x, int y, int s) { for (int j = s; j <= 18; ++j) { swap(a[x][j], a[y][j]); } } void relax(int x, int y, int s) { for (int j = s; j <= 18; ++j) { a[y][j] ^= a[x][j]; } } }M; void solve(int R) { int f1 = 0, f2 = 0, x1[20], x2[20], t1, t2, ans1 = 0x3fffffff, ans2 = 0x3fffffff; for (int i = R + 1; i <= 16; ++i) { if (M.a[i][17]) f1 = 1; if (M.a[i][18]) f2 = 1; } if (f1 && f2) { puts("Impossible"); return; } for (int s = 0; s < 16; ++s) { t1 = t2 = 0; memset(x1, 0, sizeof (x1)); memset(x2, 0, sizeof (x2)); for (int i = 0; i < 4; ++i) { x1[R+i+1] = s & (1 << i) ? 1 : 0; x2[R+i+1] = x1[R+i+1]; if (x1[R+i+1]) ++t1; if (x2[R+i+1]) ++t2; } for (int i = R; i >= 1; --i) { for (int j = i + 1; j <= 16; ++j) { if (!f1) x1[i] ^= (x1[j] * M.a[i][j]); if (!f2) x2[i] ^= (x2[j] * M.a[i][j]); } if (!f1) { x1[i] ^= M.a[i][17]; if (x1[i]) ++t1; } if (!f2) { x2[i] ^= M.a[i][18]; if (x2[i]) ++t2; } } if (!f1) ans1 = min(ans1, t1); if (!f2) ans2 = min(ans2, t2); } printf("%d\n", min(ans1, ans2)); } void Gauss() { int i = 1, k; for (int j = 1; j <= 16; ++j) { // 枚举上三角矩阵的对角线 for (k = i; k <= 16; ++k) { if (M.a[k][j]) break; } if (k > 16) continue; if (k != i) { // 如果该行就是第首行,那么我们直接进行消元,否则交换行,使得第一行为单位1 M.rswap(k, i, j); } for (k = i + 1; k <= 16; ++k) { // 从下一行开始进行消元 if (M.a[k][j]) { M.relax(i, k, j); } } ++i; } solve(i - 1); } int main() { M.init(); for (int i = 1; i <= 4; ++i) { scanf("%s", G[i] + 1); } for (int i = 1; i <= 4; ++i) { for (int j = 1; j <= 4; ++j) { M.a[TO(i, j)][17] = G[i][j] == 'w'; M.a[TO(i, j)][18] = G[i][j] == 'b'; } } Gauss(); return 0; }