状态压缩搜索——【USACO2.2.4】派对灯

一开始看还以为状态100种，搜索深度10000层，明显要爆内存或时间

看了nocow解释，每种状态里每六个灯循环，就可以将基础状态找出：（按钮4：当按下此按钮，将改变所有序号是3*K+1(K>=0)的灯。例如：1,4,7... ）这里可以发现循环

一共就8种

bool map[8][6]={

{0,0,0,0,0,0},

{0,0,1,1,1,0},

{0,1,0,1,0,1},

{0,1,1,0,1,1},

{1,0,0,1,0,0},

{1,0,1,0,1,0},

{1,1,0,0,0,1},

{1,1,1,1,1,1}};

bu[8]={1,2,1,1,2,1,2,0};//达到每种状态的步数

c==0 搜状态

c==1 搜

c>=1 搜

然后就是要注意没有可能的状态，则输出一行'IMPOSSIBLE'就行了……

#include<stdio.h>

bool map[8][6]=
{
    {0,0,0,0,0,0},
    {0,0,1,1,1,0},
    {0,1,0,1,0,1},
    {0,1,1,0,1,1},
    {1,0,0,1,0,0},
    {1,0,1,0,1,0},
    {1,1,0,0,0,1},
    {1,1,1,1,1,1}
};



int bu[8]={1,2,1,1,2,1,2,0};
bool kai[6];
bool guan[6];

int main()
{
    int temp,n,c,add;
    while(scanf("%d",&n)!=EOF)
    {
        int i,j;
        for(i=0;i<6;i++)
        {
            kai[i]=0;
            guan[i]=0;
        }
        scanf("%d",&c);
        while(scanf("%d",&temp),temp!=-1)
        {
            kai[(temp-1)%6]=1;
        }
        while(scanf("%d",&temp),temp!=-1)
        {
            guan[(temp-1)%6]=1;
        }
        
        bool one=0,hou=0;
        if(c==0)
        {
            for(i=0;i<6;i++)
            {
                if(guan[i]==1)
                {
                    printf("IMPOSSIBLE\n");
                    one=1;
                    break;
                }
            }
            
            if(one==0)
            {
                for(i=0;i<n;i++)
                {
                    printf("1");
                }
                printf("\n");
            }
        }
        else if(c==1)
        {
            for(i=0;i<8;i++)
            {
                if(bu[i]>1) continue;
                
                add=0;
                for(j=0;j<6;j++)
                {
                    if(kai[j]==1)
                    {
                        if(map[i][j]==1)
                        {}
                        else
                            break;
                    }
                    if(guan[j]==1)
                    {
                        if(map[i][j]==0)
                        {}
                        else
                            break;
                    }
                    add++;
                }
                int k;
                if(add==6)
                {
                    hou=1;
                    for(k=0;k<n;k++)
                        printf("%d",map[i][k%6]);
                    printf("\n");
                }
            }
             if(hou==0)
                printf("IMPOSSIBLE\n");
        }
        else
        {
            for(i=0;i<8;i++)
            {
                
                add=0;
                for(j=0;j<6;j++)
                {
                    if(kai[j]==1)
                    {
                        if(map[i][j]==1)
                        {}
                        else
                            break;
                    }
                    if(guan[j]==1)
                    {
                        if(map[i][j]==0)
                        {}
                        else
                            break;
                    }
                    add++;
                }
                int k;
                if(add==6)
                {
                    hou=1;
                    for(k=0;k<n;k++)
                        printf("%d",map[i][k%6]);
                    printf("\n");
                    
                }
            }
            if(hou==0)
                printf("IMPOSSIBLE\n");
        }
    }
}