POJ 1222 高斯消元

来源：互联网发布：概率矩阵分解中心极限编辑：程序博客网时间：2024/05/21 10:19

解题思路：

30个格子，对每一个格子建立一个方程，高斯消元

#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#define LL lonA lonA#define FOR(i, x, y) for(int i=x;i<=y;i++)using namespace std;const int MAXN = 50;int A[MAXN][MAXN];int x[MAXN];void init(){    memset(A, 0, sizeof(A));    for(int i=0;i<5;i++)    {        for(int j=0;j<6;j++)        {            int p = 6 * i + j;            A[p][p] = 1;            if(i > 0) A[(i-1)*6+j][p] = 1;            if(i < 4) A[(i+1)*6+j][p] = 1;            if(j > 0) A[i*6+j-1][p] = 1;            if(j < 5) A[i*6+j+1][p] = 1;        }    }}void Gauss(){    int k;    int row, col;    for (row =0, col =0; row <30&& col <30; row++, col++)    {        for (k = row; k <30; k++)            if (A[k][col] !=0)                break;        if (k ==30)        {            row--;            continue;        }        if (k != row)            for (int i = col; i <=30; i++)                swap(A[row][i], A[k][i]);        for (int i = row +1; i <30; i++)            if (A[i][col])                for (int j = col; j <=30; j++)                    A[i][j] ^= A[row][j];    }    for (int i = row; i >=0; i--)    {        x[i] = A[i][30];        for (int j =29; j > i; j--)            x[i] ^= (A[i][j] && x[j]);    }}int main(){    int T, kcase = 1;    scanf("%d", &T);    while(T--)    {        init();        for(int i=0;i<30;i++)            scanf("%d", &A[i][30]);        Gauss();        printf("PUZZLE #%d\n", kcase++);        for(int i=0;i<30;i++)        {            if((i + 1) % 6 == 0) printf("%d\n", x[i]);            else printf("%d ", x[i]);        }    }    return 0;}

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