zoukankan      html  css  js  c++  java
  • Codeforces Gym 100650C The Game of Efil DFS

    The Game of Efil
    Time Limit: 20 Sec

    Memory Limit: 256 MB

    题目连接

    http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88443#problem/C

    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 generation. • 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 3 2 0 0 0 1 3 3 4 0 0 0 1 0 2 1 1 3 3 5 0 0 1 0 1 2 2 1 2 2 0 0

    Sample Output

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

    HINT

    题意

    给你一个病毒产生或者消亡的规律,然后让你求这个状态下的图的上一状态一共有多少种

    题解

    数据范围很小,n*m<=16,所以直接dfs出所有的状态就好了

    然后再check

    代码:

    //qscqesze
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <ctime>
    #include <iostream>
    #include <algorithm>
    #include <set>
    #include <vector>
    #include <sstream>
    #include <queue>
    #include <typeinfo>
    #include <fstream>
    #include <map>
    #include <stack>
    typedef long long ll;
    using namespace std;
    //freopen("D.in","r",stdin);
    //freopen("D.out","w",stdout);
    #define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
    #define maxn 200051
    #define mod 10007
    #define eps 1e-9
    int Num;
    //const int inf=0x7fffffff;   //нчоч╢С
    const int inf=0x3f3f3f3f;
    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;
    }
    //**************************************************************************************
    int M[30][30];
    int d[30][30];
    int cnt[30][30];
    int d2[30][30];
    int n,m;
    int ans=0;
    int dx[8]={1,-1,1,-1,1,-1,0,0};
    int dy[8]={1,-1,-1,1,0,0,1,-1};
    int C(int x,int k)
    {
        if(x==-1)
            x=k-1;
        if(x==k)
            return 0;
        return x;
    }
    int check()
    {
        memset(cnt,0,sizeof(cnt));
        memset(d2,0,sizeof(d2));
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                for(int k=0;k<8;k++)
                {
                    int xx=C(i+dx[k],n);
                    int yy=C(j+dy[k],m);
                    if(d[xx][yy]==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;
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(d2[i][j]!=M[i][j])
                    return 0;
            }
        }
        return 1;
    }
    int tot=0;
    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 t=1;
        while(cin>>n>>m)
        {
            if(n==0&&m==0)
                break;
            memset(M,0,sizeof(M));
            ans=0;
            int k=read();
            for(int i=0;i<k;i++)
            {
                int x=read(),y=read();
                M[x][y]=1;
            }
            dfs(0,0);
            if(ans!=0)
                printf("Case %d: %d possible ancestors.
    ",t++,ans);
            else
                printf("Case %d: Garden of Eden.
    ",t++);
        }
    }
  • 相关阅读:
    使用pandas的get_dummies对类目型的特征因子化
    关于RandomForestRegressor,补全null数值
    关于train_test_split和cross_val_score交叉检验
    关于seaborn
    正态分布
    单下划线或双下划线的意义
    MFC中关于运行时类信息及动态创建对象的两个宏的意义(转)
    DPDK
    根据结构体成员地址得到结构体入口地址,内核代码
    多线程频繁写全局变量导致性能降低
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4740476.html
Copyright © 2011-2022 走看看