Codeforces C The Game of Efil （暴力枚举状态）

http://codeforces.com/gym/100650

阅读题，边界的cell的邻居要当成一个环形的来算，时间有8s，状态最多2^16种，所以直接暴力枚举就行了。另外一种做法是逆推。

#include<bits/stdc++.h>
using namespace std;
 
int m,n;
const int maxn = ;
int g[maxn][maxn];
int cnt[maxn][maxn];
int dx[] = {-,,-,-, ,,, };
int dy[] = { ,,-, ,-,,,-};
 
typedef pair<int,int> pii;
#define fi first
#define se second
#define PB push_back
#define MP make_pair
int tar;
 
bool check(int mask)
{
    for(int i = ; i < m; i++)
    for(int j = ; j < n; j++){
        g[i][j] = (mask>>(i*n+j))&;
    }
    vector<pii> die,birth;
    for(int i = ; i < m; i++)
    for(int j = ; j < n; j++){
        cnt[i][j] = ;
        for(int k = ; k < ; k++){
            int nx = i+dx[k],ny = j+dy[k];
            if(nx>=m) nx = ;
            if(nx<) nx = m-;
            if(ny<) ny = n-;
            if(ny>=n) ny = ;
            cnt[i][j] += g[nx][ny];
        }
        if(cnt[i][j]<||cnt[i][j]>=) die.PB(MP(i,j));
        if(!g[i][j] && cnt[i][j] == ) birth.PB(MP(i,j));
    }
    for(int i = ; i < die.size(); i++) {
        pii &t = die[i];
        g[t.fi][t.se] = ;
    }
    for(int i = ; i < birth.size(); i++){
        pii &t = birth[i];
        g[t.fi][t.se] = ;
    }
    int sta = ;
    for(int i = ; i < m; i++){
        for(int j = ; j < n; j++){
            if(g[i][j]) sta |= <<(i*n+j);
        }
    }
    return sta == tar;
}
 
int main()
{
   // freopen("in.txt","r",stdin);
    int kas = ;
    while(~scanf("%d%d",&m,&n)&&m){
        if(kas++) putchar('\n');
        int N; scanf("%d",&N);
        tar = ;
        while(N--){
            int r,c;
            scanf("%d%d",&r,&c);
            tar |= <<(r*n+c);
        }
        int M = <<(n*m);
        int ans = ;
        for(int mask = ; mask < M; mask++){
            ans += check(mask);
        }
        printf("Case %d: ",kas);
        if(ans) printf("%d possible ancestors.",ans);
        else printf("Garden of Eden.");
    }
    return ;
}