POJ 1830 (高斯消元)

又一个高斯消元的水题,求出自由元的个数n,答案就是2^n.

#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
using namespace std;
#define maxn 333
#define free Free

int a[maxn][maxn];
int n, m;
int free[maxn*maxn]; //标记自由元
int x[maxn*maxn], cnt; //解集 自由元个数
int equ, var; //方程个数 未知数个数

int Gauss () {
    int max_r, col, k; //最大的数在的行 当前处理的列
    cnt = 0; //自由元个数
    for (k = 0, col = 0; k < equ && col < var; k++, col++) {
        max_r = k;
        for (int i = k+1; i < equ; i++) { //找到最大的数所在的行
            if (abs (a[i][col]) > abs (a[max_r][col]))
                max_r = i;
        }
        if (max_r != k) { //交换最大的数所在的行和当前行
            for (int i = 0; i <= var; i++)
                swap (a[k][i], a[max_r][i]);
        }
        if (a[k][col] == 0) { //判断自由元 处理当前行的下一列
            k--;
            free[cnt++] = col;
            continue;
        }
        for (int i = k+1; i < equ; i++) {
            if (a[i][col]) {
                for (int j = col; j <= var; j++) {
                    a[i][j] ^= a[k][j];
                }
            }
        }
    }

    for (int i = k; i < equ; i++) { //判断无解情况
        if (a[i][col])
            return -1;
    }
    if (k < var)
        return var - k; //返回自由元个数

    for (int i = var-1; i >= 0; i--) { //回代
        x[i] = a[i][var];
        for (int j = i+1; j < var; j++)
            x[i] ^= (a[i][j] && x[j]);
    }
    return 0;
}

int s1[maxn], s2[maxn];

int main () {
    //freopen ("in", "r", stdin);
    int t, kase = 0;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%d", &n);
        memset (a, 0, sizeof a);
        for (int i = 0; i < n; i++) {
            scanf ("%d", &s1[i]);
        }
        for (int i = 0; i < n; i++) {
            scanf ("%d", &s2[i]);
        }
        for (int i = 0; i < n; i++) {
            if (s1[i]^s2[i]) {
                a[i][n] = 1;
            }
            else a[i][n] = 0;
        }
        for (int i = 0; i < n; i++)
            a[i][i] = 1;

        while (1) {
            int u, v;
            scanf ("%d%d", &u, &v);
            if (u == 0 && v == 0)
                break;
            v--, u--;
            a[v][u] = 1;
        }

        equ = n, var = n;
        int ans = Gauss ();
        if (ans == -1)
            printf ("Oh,it's impossible~!!\n");
        else printf ("%lld\n", (long long) pow (2, ans));
    }
    return 0;
}