HDU1536&&POJ2960 S-Nim(SG函数博弈)

S-Nim

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

经典的Nim游戏题目中已经给出了——每一堆选取的数量没有限制。

S-Nim游戏仅仅是限制了每一次从每一堆中选取的个数，依旧用sg函数计算即可。

经典的Nim游戏中sg(x) = x，所以结果就是每一堆的状态直接xor即可，S-Nim游戏先计算每一堆的sg函数值，然后判断方法依旧是用xor。

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 102
int n,m,l,s[N],sg[10005],ans,a;
bool vis[10005];
void c()
{
    sg[0]=0;
    for(int i=1;i<=10000;i++)
    {
        memset(vis,0,sizeof(vis));
        for(int j=0;j<n&&s[j]<=i;j++)
        vis[sg[i-s[j]]]=1;
        for(int x=0;x<=10000;x++)
            if(!vis[x])
        {
            sg[i]=x;
            break;
        }
    }
}
int main()
{
    while(cin>>n&&n)
    {
        for(int i=0;i<n;i++)
            cin>>s[i];
        sort(s,s+n);
        c();
        cin>>m;
        while(m--)
        {
            ans=0;
            cin>>l;
            while(l--)
            {
                cin>>a;
                ans^=sg[a];
            }
            if(ans)
                cout<<"W";
            else
                cout<<"L";
        }
        cout<<endl;
    }
    return 0;
}

 对sg函数的理解：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 102
int n,m,l,s[N],sg[10005],ans,a;
bool vis[10005];
void c()
{
    sg[0]=0;
    for(int i=1;i<=10000;i++)
    {
        memset(vis,0,sizeof(vis));
        for(int j=0;j<n&&s[j]<=i;j++)
        vis[sg[i-s[j]]]=1; //里面的sg[]是sg[i]的后继 vis[sg[]] 指的是求mex
        for(int x=0;x<=10000;x++)
            if(!vis[x])
        {
            sg[i]=x;
            break;
        }
    }
}
int main()
{
    while(cin>>n&&n)
    {
        for(int i=0;i<n;i++)
            cin>>s[i];
        sort(s,s+n);
        c();
        for(int i=0;i<30;i++)
        cout<<"i: "<<i<<"  sg[i]: "<<sg[i]<<endl;
        /*cin>>m;
        while(m--)
        {
            ans=0;
            cin>>l;
            while(l--)
            {
                cin>>a;
                ans^=sg[a];
            }
            if(ans)
                cout<<"W";
            else
                cout<<"L";
        }
        cout<<endl;*/
    }
    return 0;
}

让我们再来考虑一下顶点的SG值的意义。当g(x)=k时，表明对于任意一个0<=i<k，都存在x的一个后继y满足g(y)=i。

也就是说，当某枚棋子的SG值是k时，我们可以把它变成0、变成1、……、变成k-1，但绝对不能保持k不变。

不知道你能不能根据这个联想到Nim游戏，Nim 游戏的规则就是：每次选择一堆数量为k的石子，

可以把它变成0、变成1、……、变成k-1，但绝对不能保持k不变。这表明，如果将n枚棋子所在的顶点的 SG值看作n

堆相应数量的石子，那么这个Nim游戏的每个必胜策略都对应于原来这n枚棋子的必胜策略！