soj1171 The Game of Efil

Constraints

Time Limit: 2 secs, Memory Limit: 64 MB

Description

Almost anyone who has ever taken a class in computer science is familiar with the "Game of Life," John Conway's cellular automata with extremely simple rules of birth, survival, and death that can give rise to astonishing complexity.
The game is played on a rectangular field of cells, each of which has eight neighbors (adjacent cells). A cell is either occupied or not. The rules for deriving a generation from the previous one are:

• If an occupied cell has 0, 1, 4, 5, 6, 7, or 8 occupied neighbors, the organism dies (0, 1: of loneliness; 4 thru 8: of overcrowding).
• If an occupied cell has two or three occupied neighbors, the organism survives to the next gener-ation.
• If an unoccupied cell has three occupied neighbors, it becomes occupied (a birth occurs).

One of the major problems researchers have looked at over the years is the existence of so-called "Garden of Eden" configurations in the Game of Life -- configurations that could not have arisen as the result of the application of the rules to some previous configuration. We're going to extend this question, which we'll call the "Game of Efil": Given a starting configuration, how many possible parent configurations could it have? To make matters easier, we assume a finite grid in which edge and corner cells "wrap around" (i.e., a toroidal surface). For instance, the 2 by 3 configuration:

has exactly three possible parent configurations; they are:

You should note that when counting neighbors of a cell, another cell may be counted as a neighbor more than once, if it touches the given cell on more than one side due to the wrap around. This is the case for the configurations above.

Input

There will be multiple test cases. Each case will start with a line containing a pair of positive integers m and n, indicating the number of rows and columns of the configuration, respectively. The next line will contain a nonnegative integer k indicating the number of "live" cells in the configuration. The following k lines each contain the row and column number of one live cell, where row and column numbering both start at zero. The final test case is followed by a line where m = n = 0 -- this line should not be processed. You may assume that the product of m and n is no more than 16.

Output

For each test case you should print one line of output containing the case number and the number of possible ancestors. Imitate the sample output below. Note that if there are 0 ancestors, you should print out
Garden of Eden.

Sample Input

2 320 00 13 340 00 10 21 13 350 01 01 22 12 20 0

Sample Output

Case 1: 3 possible ancestors.Case 2: 1 possible ancestors.Case 3: Garden of Eden.

题目大意：

生命游戏相关的题目，周围的邻居数目决定了下一代的个体存活与否，注意一下有限的地图内是wrap around的，即卷在一起的曲面，即左右其实相邻，上下也相邻，还有四个角也相邻，刚开始还以为这个面板是可以旋转的，旋转相等也算，结果似乎我想复杂了。问当前状态有多少种可能的父代会得到它。

解题方法：

额因为面板很小，最多16个格子而已，所以枚举父代格子的所有情况，一一判断是否会生成一个与输入相同的儿子就好了，枚举用dfs记录所有格子的状态。注意这里要上下左右相连，那么就重新定义一个大一圈的地图，将最右边一列赋给0列，将第一列赋给m+1列，上下同理，别忘了还有四个角，然后就对里面的每一个点按游戏规则进行判断存活与否就好了

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int n,m;int map[20][20];int mm[20][20];int ans[20][20];int ansnum;void dfs(int now){if (now==n*m){memset(map,0,sizeof(map));//memset(tmp,0)for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){map[i][j]=mm[i-1][j-1];//cout<<map[i][j];}//cout<<endl;}//上下左右的连接for(int i=1;i<=m;i++){map[0][i]=map[n][i];}for(int i=1;i<=m;i++){map[n+1][i]=map[1][i];} for (int i=1;i<=n;i++){map[i][0]=map[i][m];}for (int i=1;i<=n;i++){map[i][m+1]=map[i][1];}//四个角map[0][0]=map[n][m];map[0][m+1]=map[n][1];map[n+1][0]=map[1][m];map[n+1][m+1]=map[1][1];int tmp[20][20]={0};for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){int sum=map[i-1][j-1]+map[i][j-1]+map[i+1][j-1]+map[i-1][j]+map[i+1][j]+map[i-1][j+1]+map[i][j+1]+map[i+1][j+1];if (map[i][j]==1){if (sum==3||sum==2){tmp[i][j]=1;}else tmp[i][j]=0;}else if(sum==3) tmp[i][j]=1;}}for (int i=1;i<=n;i++){for (int j=1;j<=m;j++){if (tmp[i][j]!=ans[i][j]){return ;}}}ansnum++;return;}//深搜枚举所有情况int row=now/m;int col=now%m;mm[row][col]=1;dfs(now+1);mm[row][col]=0;dfs(now+1); } int main(){int co=1;while(cin>>n>>m){if (n==0) break;int num;cin>>num;ansnum=0;int x,y;memset(ans,0,sizeof(ans));memset(mm,0,sizeof(mm));for (int i=0;i<num;i++){cin>>x>>y;ans[x+1][y+1]=1;}dfs(0);printf("Case %d: ",co++);if (ansnum==0){printf("Garden of Eden.\n");}else {printf("%d possible ancestors.\n",ansnum);}}}