POJ1753 Flip Game 高斯消元

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;
}