vvalive 6323 상태 압축 DP

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<cstring>

using namespace std;

int dp[5000][13][40],g[5][5],d[13][13],n,l,ans;

bool vi[13],ntou[13];

struct Point{

    int x,y;

}p[13];

bool OK(int a,int b)

{

    int x1,y1,x2,y2,i;

    x1=p[a].x,y1=p[a].y;

    x2=p[b].x,y2=p[b].y;

    if(x1==x2){

        if(y1>y2)

            swap(y1,y2);

        for(i=y1+1;i<y2;i++) if(!vi[g[x1][i]]||ntou[g[x1][i]])

        return false;

    }

    if(y1==y2){

        if(x1>x2)

            swap(x1,x2);

        for(i=x1+1;i<x2;i++) if(!vi[g[i][y1]]||ntou[g[i][y1]])

            return false;

    }

    if(abs(x1-x2)==2&&abs(y1-y2)==2){

        int x=(x1+x2)/2;

        int y=(y1+y2)/2;

        if(!vi[g[x][y]]||ntou[g[x][y]])

            return false;

    }

    return true;

}

void init()

{

    int i,j,cnt=0;

    for(i=1;i<=3;i++){

        for(j=1;j<=4;j++){

            g[i][j]=++cnt;

            p[cnt].x=i,p[cnt].y=j;

        }

    }

    for(i=1;i<=12;i++){

        for(j=1;j<=12;j++){

            d[i][j]=abs(p[i].x-p[j].x)+abs(p[i].y-p[j].y);

        }

    }

}

void solve()

{

    int i,j,k,r;

    ans=0;

    int N=(1<<12)-1;

    for(i=0;i<12;i++)

        dp[1<<i][i+1][0]=1;

    for(i=1;i<=N;i++){

        memset(vi,0,sizeof(vi));

        for(k=0;k<12;k++) if(((1<<k)&i)) vi[k+1]=1;

        for(j=0;j<12;j++) if((((1<<j)&i))&&!ntou[j+1]){

            for(k=0;k<12;k++) if(((1<<k)&i)==0&&!ntou[k+1]&&OK(j+1,k+1)){

                //cout<<p[j+1].x<<" "<<p[j+1].y<<" "<<p[k+1].x<<" "<<p[k+1].y<<endl;

                for(r=0;r<=40;r++){

                    dp[i|(1<<k)][k+1][r+d[j+1][k+1]]+=dp[i][j+1][r];

                }

            }

        }

    }

    for(i=1;i<=N;i++){

        for(j=1;j<=12;j++){

            ans+=dp[i][j][l];

        }

    }

}

int main()

{

    int i,j,t,x,y;

    init();

    scanf("%d",&t);

    while(t--){

        memset(dp,0,sizeof(dp));

        memset(vi,0,sizeof(vi));

        memset(ntou,0,sizeof(ntou));

        scanf("%d%d",&l,&n);

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

            scanf("%d%d",&x,&y);

            ntou[g[x][y]]=1;

        }

        if(l>=40){

            printf("BAD MEMORY
");

            continue;

        }

        solve();

        if(ans)

            printf("%d
",ans);

        else

            printf("BAD MEMORY
");

    }

    return 0;

}