POJ 3185 The water bowls (math: Gaussian elimination)

Each location affects only adjacent locations

Do the Gaussian elimination and then enumerate.

The code is as follows:

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAXN usi

NG namespace Std;
int X[MAXN];
int FREE_X[MAXN];

int A[MAXN][MAXN];
        void print () {for (Int. i=0; i<20; ++i) {for (int j=0; j<=20; ++j) printf ("%d", a[i][j]);
    Puts ("");
    }} int Gauss (int equ, int var) {int I, J, K, Col;
    int TA, TB, LCM;
    int tmp, FREE_INDEX;

    int Free_x_num, Max_r;
    int num = 0;
        for (int i=0; i<=var; ++i) {x[i] = 0;
    Free_x[i] = 0;
    } col = 0;
        for (k=0; k<equ&&col<var; ++k, ++col) {max_r = k;
        for (i=k+1; i<equ; ++i) {if (ABS (A[i][col]) > abs (A[max_r][col])) Max_r = i;
        } if (Max_r! = k) for (j=k; j<=var; ++j) Swap (A[k][j], a[max_r][j]);
            if (a[k][col] = = 0) {--k;
            free_x[num++] = col;
        Continue
    }    For (i=k+1, i<equ; ++i) {if (A[i][col]) {for (j=col; j<=var; ++j)
            A[I][J] ^= a[k][j];
    }}} for (i=k; i<equ; ++i) {if (A[i][col]) return-1;
} return var-k;
    } void Solve (int equ, int var) {int ans = Gauss (equ, Var);
        if (ans = = 0) {int res = 0;
        for (int i=0; i<equ; ++i) Res + = X[i];
        printf ("%d\n", res);
    return;
        } else {int res = 0x3f3f3f3f;
        int tot = (1<<ans);
            for (int i=0; i<tot; ++i) {int cnt = 0;
                    for (int j=0; j<ans; ++j) {if (i& (1<<j)) {X[free_x[j]] = 1; 
                ++cnt;
            } else X[free_x[j]] = 0;
                } for (int j=var-ans-1; j>=0;--j) {int tmp = A[j][var];
                   for (int l=j+1; l<var; ++l) if (A[j][l])     TMP ^= X[L];
                X[J] = tmp;
            if (X[j]) ++cnt;
        } res = min (res, CNT);
    } printf ("%d\n", res);
    }} int main (void) {int nx;
        while (~SCANF ("%d", &nx)) {memset (A, 0, sizeof (a));
        A[0][20] = NX;

        for (int i=1; i<20; ++i) scanf ("%d", &a[i][20]);
        A[0][0] = a[0][1] = 1;
        A[19][18] = a[19][19] = 1;
        for (int i=1; i<19; ++i) {a[i][i] = a[i][i-1] = a[i][i+1] = 1;
        }//print ();
        Solve (20, 20);
    Print ();
} return 0; }