Codeforces Gym 100650C The Game of Efil DFS

题目链接:

http://codeforces.com/gym/100650/attachments

题意:

每个细胞如果四周细胞太少了，就会孤独而死，如果细胞周围细胞太多了，就会挤死。给你个布局，问你他的上一代布局会有几种。不过这道题的关键在于wrap around这个词，即边界是循环的。

题解:

数据范围很小,n*m<=16，所以直接dfs出所有的状态
然后再check

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e5+10;

int n,m,k,ans;
int G[20][20],d[20][20],d2[20][20],cnt[20][20];
int dx[8] = {0,0,1,-1,1,1,-1,-1};
int dy[8] = {1,-1,0,0,1,-1,1,-1};

int C(int x,int k){
    if(x == -1) return k-1;
    if(x == k) return 0;
    return x;
}

bool check(){
    MS(d2); MS(cnt);
    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++){
            for(int k=0; k<8; k++){
                int x = C(i+dx[k],n);
                int y = C(j+dy[k],m);
                if(d[x][y] == 1)
                    cnt[i][j]++;
            }
        }

    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++){
            if(d[i][j] == 1){
                if(cnt[i][j]==2 || cnt[i][j]==3)
                    d2[i][j] = 1;
                else
                    d2[i][j] = 0;
            }else{
                if(cnt[i][j] == 3)
                    d2[i][j] = 1;
                else
                    d2[i][j] = 0;
            }
        }

    for(int i=0; i<n; i++)
        for(int j=0; j<m; j++)
            if(G[i][j] != d2[i][j])
                return 0;

    return 1;
}

void dfs(int x,int y){
    if(x == n){
        if(check()) ans++;
        return ;
    }

    d[x][y] = 0;
    if(y == m-1)
        dfs(x+1,0);
    else
        dfs(x,y+1);
    d[x][y] = 1;
    if(y == m-1)
        dfs(x+1,0);
    else
        dfs(x,y+1);
    d[x][y] = 0;
}

int main(){
    int cas=1;
    while(cin>>n>>m,(n+m)){
        MS(d); MS(G); ans=0;

        cin >> k;
        for(int i=0; i<k; i++){
            int x,y; cin>>x>>y;
            G[x][y] = 1;
        }
        dfs(0,0);
        if(ans != 0)
            cout << "Case " << cas++ << ": " << ans << " possible ancestors.
";
        else
            cout << "Case " << cas++ << ": Garden of Eden.
";
    }

    return 0;
}

// http://codeforces.com/gym/100650/attachments