POJ-1753 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;

}