POJ 1830 开关问题

Posted 2020-12-03 evenbao

【题目链接】

            http://poj.org/problem?id=1830

【算法】

           列出异或方程组，用高斯消元求解

【代码】

         

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h>
using namespace std;

int i,j,k,T,n,x,y,ans;
int a[100];

int main() 
{
        
        scanf("%d",&T);
        while (T--)
        {
                scanf("%d",&n);
                for (i = 1; i <= n; i++) scanf("%d",&a[i]);
                for (i = 1; i <= n; i++)
                {
                        scanf("%d",&x);
                        a[i] ^= x;
                        a[i] |= (1 << i);    
                }        
                while (scanf("%d%d",&x,&y) && x && y) a[y] |= (1 << x);
                ans = 1;
                for (i = 1; i <= n; i++)
                {
                        for (j = i + 1; j <= n; j++)
                        {
                                if (a[j] > a[i])
                                        swap(a[i],a[j]);
                        }
                        if (a[i] == 0)
                        {
                                ans = 1 << (n - i + 1);
                                break;
                        }
                        if (a[i] == 1)
                        {
                                ans = 0;
                                break;
                        }
                        for (k = n; k; k--)
                        {
                                if (a[i] & (1 << k))
                                {
                                        for (j = 1; j <= n; j++)
                                        {
                                                if (i != j && (a[j] & (1 << k)))
                                                        a[j] ^= a[i];
                                        }
                                        break;
                                }
                        }
                }
                if (!ans) printf("Oh,it‘s impossible~!!
");
                else printf("%d
",ans);
        }
        
        return 0;
    
}