[hdu3364]xor方程组消元

题意：n个灯，m个开关，给定每个开关控制的灯，全部的灯初始时全部熄灭，开关按一下其所控制的灯的状态全部反转，开关最多只能按一下。问达到目标状态的方案数。

思路：xor方程组的模型。

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/* ******************************************************************************** */
#include <iostream>                                                                 //
#include <cstdio>                                                                   //
#include <cmath>                                                                    //
#include <cstdlib>                                                                  //
#include <cstring>                                                                  //
#include <vector>                                                                   //
#include <ctime>                                                                    //
#include <deque>                                                                    //
#include <queue>                                                                    //
#include <algorithm>                                                                //
#include <map>                                                                      //
#include <cmath>                                                                    //
using namespace std;                                                                //
                                                                                    //
#define pb push_back                                                                //
#define mp make_pair                                                                //
#define X first                                                                     //
#define Y second                                                                    //
#define all(a) (a).begin(), (a).end()                                               //
#define fillchar(a, x) memset(a, x, sizeof(a))                                      //
                                                                                    //
typedef pair<int, int> pii;                                                         //
typedef long long ll;                                                               //
typedef unsigned long long ull;                                                     //
                                                                                    //
#ifndef ONLINE_JUDGE                                                                //
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}    //
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>                    //
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;          //
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>      //
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>              //
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>   //
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}   //
#endif // ONLINE_JUDGE                                                              //
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}        //
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}        //
template<typename T>                                                                //
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}            //
template<typename T>                                                                //
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}            //
                                                                                    //
const double PI = acos(-1.0);                                                       //
const int INF = 1e9 + 7;                                                            //
                                                                                    //
/* -------------------------------------------------------------------------------- */
 
struct XorElimination {
    const static int maxn = 55;
    bool A[maxn][maxn];
    int solve(int m, int n) {
        int i = 0, j = 0, k, r, u;
        while (i < m && j < n) {
            r = i;
            for (k = i; k < m; k ++) {
                if (A[k][j]) {
                    r = k;
                    break;
                }
            }
            if (A[r][j]) {
                if (r != i) for (k = 0; k <= n; k ++) swap(A[r][k], A[i][k]);
                for (int u = i + 1; u < m; u ++) {
                    if (A[u][j]) {
                        for (k = i; k <= n; k ++) A[u][k] ^= A[i][k];
                    }
                }
                i ++;
            }
            j ++;
        }
        /** 返回自由变元个数，无解返回-1 **/
        for (int i = 0; i < m; i ++) {
            if (A[i][n]) {
                bool ok = false;
                for (int j = 0;  j < n; j ++) {
                    if (A[i][j]) {
                        ok = true;
                        break;
                    }
                }
                if (!ok) return -1;
            }
        }
        return n - i;
    }
};/** 下标从0开始 **/
 
XorElimination solver;
bool buf[55][55];
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif // ONLINE_JUDGE
    int T, n, m, cas = 0;
    cin >> T;
    while (T --) {
        cin >> n >> m;
        memset(buf, 0, sizeof(buf));
        for (int i = 0; i < m; i ++) {
            int x, y;
            scanf("%d", &x);
            for (int j = 0; j < x; j ++) {
                scanf("%d", &y);
                buf[y - 1][i] = true;
            }
        }
        printf("Case %d:\n", ++ cas);
        int q;
        cin >> q;
        while (q --) {
            for (int i = 0; i < n; i ++) {
                for (int j = 0; j < m; j ++) {
                    solver.A[i][j] = buf[i][j];
                }
            }
            for (int i = 0; i < n; i ++) {
                int x;
                scanf("%d", &x);
                solver.A[i][m] = x;
            }
            int r = solver.solve(n, m);
            cout << (~r? (1ll << r) : 0) << endl;
        }
    }
    return 0;
}
/* ******************************************************************************** */

 

posted @ 2015-08-03 01:05  jklongint  阅读(433)  评论(0)    收藏  举报