hdu3364 Lanterns （高斯消元）

最新推荐文章于 2021-01-03 16:35:55 发布

fanlinqiang

最新推荐文章于 2021-01-03 16:35:55 发布

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Lanterns

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 10 Accepted Submission(s) : 8

Problem Description

Alice has received a beautiful present from Bob. The present contains n lanterns and m switches. Each switch controls some lanterns and pushing the switch will change the state of all lanterns it controls from off to on or from on to off. A lantern may be controlled by many switches. At the beginning, all the lanterns are off.

Alice wants to change the state of the lanterns to some specific configurations and she knows that pushing a switch more than once is pointless. Help Alice to find out the number of ways she can achieve the goal. Two ways are different if and only if the sets (including the empty set) of the switches been pushed are different.

Input

The first line contains an integer T (T<=5) indicating the number of test cases.
The first line of each test case contains an integer n (1<=n<=50) and m (1<=m<=50).
Then m lines follow. Each line contains an integer k (k<=n) indicating the number of lanterns this switch controls.
Then k integers follow between 1 and n inclusive indicating the lantern controlled by this switch.
The next line contains an integer Q (1<=Q<=1000) represent the number of queries of this test case.
Q lines follows. Each line contains n integers and the i-th integer indicating that the state (1 for on and 0 for off) of the i-th lantern of this query.

Output

For each test case, print the case number in the first line. Then output one line containing the answer for each query.
Please follow the format of the sample output.

Sample Input

Sample Output

Case 1:
1
0
Case 2:
8
0

Source

“光庭杯”第五届华中北区程序设计邀请赛暨 WHU第八届程序设计竞赛

题意：n盏灯，m个开关，每个开关控制一些灯。给出n个灯的状态，询问开关的方案数，使得达到给出的状态。

灯的状态是由控制它的所有开关的状态异或而来，所以每个开关不是0，就是1。

高斯消元后，等式左边系数全为0，等式右边系数不为0，显然无解。否则答案就是2^(自由元的个数)。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

bool ctr[55][55],fuc[55][55];
int equ,var;

__int64 Guass()
{
    int row=0,col=0;
    for( int i,j; row<equ && col<var; row++,col++ )
    {
        for(i=row;i<equ;i++)
            if(fuc[i][col]) break;

        if(i==equ)
        {
            row--;
            continue;
        }

        if(i!=row)
        {
            for(j=col;j<=var;j++)
                swap(fuc[row][j],fuc[i][j]);
        }

        for(i=row+1;i<equ;i++)
        {
            if(fuc[i][col])
            {
                for(j=col;j<=var;j++)//异或消元
                    fuc[i][j]^=fuc[row][j];
            }
        }

    }

    for(int i=row;i<equ;i++)
        if(fuc[i][var])
            return 0;
    return 1LL << (var-row);//解为 2^(m-row)
}
int main()
{
    int cas,k,m,tmp,Q;
    scanf("%d",&cas);
    for(k=1;k<=cas;k++)
    {
        printf("Case %d:\n",k);
        scanf("%d%d",&equ,&var);
        memset(ctr,0,sizeof(ctr));
        for(int i=0;i<var;i++)
        {
            scanf("%d",&m);
            while(m--)
            {
                scanf("%d",&tmp);
                ctr[tmp-1][i]=1;
            }
        }
        scanf("%d",&Q);
        while(Q--)
        {
            for(int i=0;i<equ;i++)
                scanf("%d",&fuc[i][var]);

            for(int i=0;i<equ;i++)
            {
                for(int j=0;j<var;j++)
                    fuc[i][j]=ctr[i][j];
            }
            printf("%I64d\n",Guass());
        }
    }
    return 0;
}