uva10001 - Garden of Eden

题意不好理解，参考了人家的代码才看懂的题意，

就是用某种编译方式，检测所给字符串是否合理，即求它的原串。两者长度一样，只不过新得的串是根据所给编译方式和左邻右舍的值来求得的。

对于编译方式 rule：0

其意思为:

0 0 0 -->0

0 0 1 -->0

0 1 0 -->0

0 1 1 -->0

......

通俗来讲就是对于编译方式k，先把k转化为八位二进制串，8 -->0 0 0 0 1 0 0 0.。 用八位二进制串中的每个字符分别对应{0,0,0},{0,0,1},{0,1,0},{0,1,1},{1,0,0},{1,0,1},{1,1,0},{1,1,1}

跟详细的解释请见于：http://mathworld.wolfram.com/ElementaryCellularAutomaton.html

代码如下：

#include <cstdio>
int ok, nosolve[35], gain[35], state[8], array[8][3] = {{0,0,0},{0,0,1},{0,1,0},{0,1,1},
                                    {1,0,0},{1,0,1},{1,1,0},{1,1,1}};
void to_binary(int k)
{
    for(int i = 0; i < 8; i++) {state[i] = k%2; k/=2;}
}
void dfs(int n, int cur)
{
    if(ok) return;
    if(cur==n)
    {
        if(gain[n]==gain[0]&&gain[n+1]==gain[1]) ok = 1;
        return;
    }
    for(int i = 0; i < 8; i++)
    {
        if(nosolve[cur]==state[i]&&(!cur||(gain[cur]==array[i][0]&&gain[cur+1]==array[i][1])))
        {
            if(!cur) {gain[0] = array[i][0]; gain[1] = array[i][1];}
            gain[cur+2] = array[i][2];
            dfs(n, cur+1);
        }
    }
}
int main ()
{
    int k, n;
    char s[35];
    while(scanf("%d%d%s",&k,&n,s)==3)
    {
        for(int i = 0; i < n; i++) nosolve[i] = s[i]-'0';
        to_binary(k);
        ok = 0;
        dfs(n, 0);
        if(ok)puts("REACHABLE");
        else puts("GARDEN OF EDEN");
    }
    return 0;
}

再贴一下，优化之后的代码。跑了132ms。

#include <cstdio>
int ok, nosolve[35], gain[35], state[8], array[8][3] = {{0,0,0},{0,0,1},{0,1,0},{0,1,1},
                                    {1,0,0},{1,0,1},{1,1,0},{1,1,1}};
void to_binary(int k)
{
    for(int i = 0; i < 8; i++) {state[i] = k%2; k/=2;}
}
void dfs(int n, int cur)
{
    if(ok) return;
    if(cur==n)
    {
        if(gain[n]==gain[0]&&gain[n+1]==gain[1]) ok = 1;
        return;
    }
    if(!cur) {for(int i = 0; i < 8; i++) if(nosolve[0]==state[i])
    { gain[0] = array[i][0]; gain[1] = array[i][1]; gain[2] = array[i][2]; dfs(n,cur+1); }}
    else
    {
        int t = gain[cur]*4+gain[cur+1]*2;
        if(state[t]==nosolve[cur])  {gain[cur+2] = 0; dfs(n, cur+1); }
        if(state[t+1]==nosolve[cur])  {gain[cur+2] = 1; dfs(n,cur+1); }
    }
}
int main ()
{
    int k, n;
    char s[35];
    while(scanf("%d%d%s",&k,&n,s)==3)
    {
        for(int i = 0; i < n; i++) nosolve[i] = s[i]-'0';
        to_binary(k);
        ok = 0;
        dfs(n, 0);
        if(ok)puts("REACHABLE");
        else puts("GARDEN OF EDEN");
    }
    return 0;
}