Nim博弈

Nim 游戏的拓展、每次只能取限定个数个石子。 据说所有的组合游戏都可以转化为Nim游戏模型并用SG函数求解。 自己写了一遍SG函数的实现过程、感觉就是一个记忆化搜索的过程…… S-Nim Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3506    Accepted Submission(s): 1539   Problem Description Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:   The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.   The players take turns chosing a heap and removing a positive number of beads from it.   The first player not able to make a move, loses. Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:   Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).   If the xor-sum is 0, too bad, you will lose.   Otherwise, move such that the xor-sum becomes 0. This is always possible. It is quite easy to convince oneself that this works. Consider these facts:   The player that takes the last bead wins.   After the winning player's last move the xor-sum will be 0.   The xor-sum will change after every move. Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.  Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?  your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.   Input Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0 < m ≤ 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by a 0 on a line of its own.   Output For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'. Print a newline after each test case.   Sample Input 2 2 5 3 2 5 12 3 2 4 7 4 2 3 7 12 5 1 2 3 4 5 3 2 5 12 3 2 4 7 4 2 3 7 12 0   Sample Output LWW WWL   Source Norgesmesterskapet 2004   Recommend LL   至于S-Nim 为什么能向 Nim 转化。感觉这个分析很到位： S-Nim类的游戏转化为Nim游戏，并将每个点当做一堆石子，该点的SG值当做该堆的石子数。但是为什么SG值有什么意义？为什么可以转化为石子数？只有理解了SG值的确切意义才能完全明白为什么S-Nim向Nim的转化。看一下以下例子:                                      SG=g(x)={0,1,2,3}=4      ............................................. ①                                      SG=g(x)={0,1,2,3,5}=4   ..............................................②                                         SG=g(x)={1,2,3,5}=0   .................................................③ 对比①和②可以发现，SG值似乎丢失了一些信息，②中的5值并没有通过SG值表示出来。对比②和③可以发现，③中信息似乎完全丢失了。但，是不是真的如此呢？回到博弈类题目的出发点：  一、博弈题目依据什么推测最终的输赢？         答：依据的是当前点处于必胜点，还是必败点。除此之外无需假借任何条件。 二、必胜点与必败点有什么性质。         答：(1) 所有终结点是必败点（P点）；                 (2) 从任何必胜点（N点）操作，至少有一种方法可以进入必败点（P点）；                 (3)无论如何操作， 从必败点（P点）都只能进入必胜点（N点）. 然后，看一下SG值，SG值为零的点，则为P点(即，必败点)，否则为N点(即，必胜点)。以上SG值的性质满足必败点与必胜点的所有性质:(1)所有终结点的SG值为零，为必败点。(2)若SG值不为零，则其为必胜点。设其值为m，则有SG=g(x)={0,1,2,....m-1}=m，大括号中的0为其后继点的SG值，即该点为必败点。所以满足从必胜点至少有一种方法进入必败点。(3)若，SG值为零的点不是终结点，则必存在一个值为m的点，使得SG=g(x)={...，m，...}，大括号中的m为其后继点的SG值，即该点为必胜点。所以满足从必败点都只能进入必胜点。          综上所诉，SG值为零的点，则为P点(即，必败点)，否则为N点(即，必胜点)。满足必胜点与必败点所有的性质。并且可以作为推测输赢的依据。例②中所谓的信息丢失并不存在，因为SG值仅仅关注零和非零，其重点在于后继点的SG值中是否存在零。例③中的信息同样并未丢失，因为其SG值为零恰恰表明了后继点中有不为零的值存在(或无后继点)，而必败点仅仅关心这一点信息。所以SG值仅有两个意义，零和非零，标明了该点为必胜点还是必败点。 因为石子的数量与SG值起到的相同的作用（标明必胜点或者必败点） ，因此S-Nim类题目转化为Nim类题目时可以将SG值与石子数量同等对待，更是因为如此S-Nim题目才可以转化为Nim类题目。    （上述分析摘自： http://www.cnblogs.com/AdaByron/archive/2011/08/07/2200981.html ）     #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int N = 100 + 10; const int M = 1000 + 10; int n; int f[N],set[N]; int mex(int v) {     int g[N]={0};     for(int i=0;i<n;i++)     {         int t=v-set[i];         if(t<0)             break;         if(f[t]==-1)             f[t]=mex(t);         g[f[t]]=1;     }     for(int i=0;i<N;i++)     {         if(!g[i])             return i;     } } int main() {     while(scanf("%d",&n)!=EOF)     {         if(n==0) break;         for(int i=0;i<n;i++)             scanf("%d",&set[i]);         sort(set,set+n);         memset(f,-1,sizeof(f));         f[0]=0;         int m;         scanf("%d",&m);         for(int i=0;i<m;i++)         {             int c,k,res=0;             scanf("%d",&c);             for(int j=0;j<c;j++)             {                 scanf("%d",&k);                 if(f[k] == -1)                     f[k]=mex(k);                 res ^= f[k];             }             res == 0 ? printf("L") : printf("W");         }         printf("\n");     }     return 0; }