SGU 260.Puzzle (异或高斯消元)

题意：

　　有n（<200）个格子，只有黑白两种颜色.可以通过操作一个格子改变它和其它一些格子的颜色。给出改变的关系和n个格子的初始颜色，输出一种操作方案使所有格子的颜色相同。

 

 

 

 

---

Solution：

　　很显然的高斯消元。

      这里采用了类似SGU275的方法做。 

/*

       解异或方程组

*/

#include <iostream>

#include <bitset>

#include <cstring>

using namespace std;

const int N = 209;

bitset<N> a[N], mask[N];

int base[N];

int Gauss ( int n )

{

    for ( int i = 1; i <= n; ++i ) {

        for ( int j = 1; j <= n; ++j ) {

            if ( a[i][j] ) {

                if ( !~base[j] ) {

                    mask[i][i] = 1;

                    base[j] = i;

                    break;

                }

                a[i] ^= a[base[j]];

                mask[i] ^= mask[base[j]];

            }

        }

    }

}



int n, m;

int main()

{

    ios::sync_with_stdio ( 0 );

    cin >> n;

    for ( int i = 1, k; i <= n; ++i ) {

        cin >> k;

        for ( int j = 1, t; j <= k; ++j ) {

            cin >> t;

            a[i][t] = 1;

        }

    }

    memset ( base, -1, sizeof base );

    Gauss ( n );

    for ( int i = 1, k; i <= n; ++i ) {

        cin >> k;

        a[n + 1][i] = k;

        a[n + 2][i] = k ^ 1;

    }

    int k;

    for ( k = 1; k <= n; ++k ) {

        if ( a[n + 1][k] == 1 ) {

            if ( !~base[k] ) {

                break;

            }

            a[n + 1] ^= a[base[k]];

            mask[n + 1] ^= mask[base[k]];

        }

    }

    if ( k > n ) {

        cout << mask[n + 1].count() << endl;

        for ( int i = 1; i <= n; ++i ) {

            if ( mask[n + 1][i] )

                cout << i << " ";

        }

    } else {

        for ( k = 1; k <= n; ++k ) {

            if ( a[n + 2][k] ) {

                if ( !~base[k] ) {

                    break;

                }

                a[n + 2] ^= a[base[k]];

                mask[n + 2] ^= mask[base[k]];

            }

        }

        if ( k > n ) {

            cout << mask[n + 2].count() << endl;

            for ( int i = 1; i <= n; ++i ) {

                if ( mask[n + 2][i] )

                    cout << i << " ";

            }

        } else {

            cout << -1 << endl;

        }

    }

}

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