SOJ 1171. The Game of Efil

解题技巧：

　　1.地图扩展expand().

　　2.目标状态压缩。

代码如下：

#include <iostream>
#include <cstdio>
using namespace std;

const int maxm = 18;
const int maxn = 18;
int m, n;

int dm[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dn[8] = {-1, 0, 1, -1, 1, -1, 0, 1};

bool altas[maxm][maxn];
int target; // 目标状态压缩
int ans;

void expand() {
    for (int i = 1; i <= n; ++i) {
        altas[0][i] = altas[m][i];
        altas[m + 1][i] = altas[1][i];
    }
    for (int i = 1; i <= m; ++i) {
        altas[i][0] = altas[i][n];
        altas[i][n + 1] = altas[i][1];
    }
    altas[0][0] = altas[m][n];
    altas[0][n + 1] = altas[m][1];
    altas[m + 1][0] = altas[1][n];
    altas[m + 1][n + 1] = altas[1][1];
}

int neighborCount(int i, int j) {
    int count = 0;
    for (int k = 0; k < 8; ++k) {
        int next_i = i + dm[k];
        int next_j = j + dn[k];
        if (altas[next_i][next_j])
            ++count;
    }
    return count;
}

bool hasLife(int i, int j) {
    int neighbors = neighborCount(i, j);
    bool hasCell = false;
    if (altas[i][j] && 2 <= neighbors && neighbors <= 3) {  // 如果原先有生命，并且邻居的数量是2-3，则存活
        hasCell = true;
    } else if (!altas[i][j] && neighbors == 3){
        hasCell = true;
    }
    return hasCell;
}

void dfs(int cur) {
    if (cur == m * n) {
        expand();
        int status = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (hasLife(i, j)) {
                    status += 1 << ((i - 1)*n + (j - 1));
                }
            }
        }
        if (status == target)
            ++ans;
        return;
    }
    int cur_m = cur / n;
    int cur_n = cur % n;
    altas[cur_m + 1][cur_n + 1] = true;
    dfs(cur + 1);
    altas[cur_m + 1][cur_n + 1] = false;
    dfs(cur + 1);
}

int main() {
    int cases = 0;
    while (++cases, scanf("%d%d", &m, &n), m != 0 && n != 0) {
        int t; scanf("%d", &t); target = 0; ans = 0;
        int ta, tb;
        while (t--) {
            scanf("%d%d", &ta, &tb);
            target += 1 << (ta*n + tb);
        }
        dfs(0);
        cout << "Case " << cases << ": ";
        ans == 0 ? (cout << "Garden of Eden." << endl) : (cout << ans << " possible ancestors." << endl);
    }
    return 0;
}