[原创]HDU 1536 S-Nim [SG函数]【博弈】

2016-07-30 16:38:04 Tabris_ 阅读数：406

博客爬取于2020-06-14 22:44:02
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52073496

题目连接 ：http://acm.hdu.edu.cn/showproblem.php?pid=1536

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S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6629 Accepted Submission(s): 2802

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

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题目大意 :
就是多个 有多个堆的取石子游戏，只有N种取法 ，问先手输或者赢。

解题思路 : 很明显的SG函数问题 通过每次给你的N种取法 把SG值求出来然后看最后异或值是否为0就行了额

！！！！！！！！！！注意的是
计算sg函数的时候 最开始我h[]数组开的是10008,这样的结果就是无限的TLE
后来发现 取法有N种 那么i-s[]最多也只有N种 那么我h[]只要开 N 这么大就完全够了 以前怎么也想不明白 但是现在想明白了。。 只有这样才不会在 预处理的时候发生超时

附本题代码
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# include<bits/stdc++.h>

using namespace std;

# define mem(a,b,c)  for(int i=0;i<(c);i++)a[i]=(b);
# define lalal puts("******");
const int M = 10008;
const int N = 108;

int s[N];
int sg[M];
int h[N];

void getSG(int t)
{
    sg[0]=0;
    for(int i=1; i<10001; i++)
    {
        memset(h,0,sizeof(h));

        for(int j=0; j<t&&i>=s[j]; j++)
            if(i-s[j]>=0)
                h[sg[i-s[j]]]=1;

        for(int j=0; j<10001; j++)
            if(!h[j])
            {
                sg[i]=j;
                break;
            }
    }
    return ;
}


int main()
{
    int k;
    while(~scanf("%d",&k)&&k)
    {
        for(int i=0; i<k; i++)
            scanf("%d",&s[i]);
        sort(s,s+k);
        getSG(k);

        int _;
        scanf("%d",&_);
        while(_--)
        {
            int n, x, ans = 0;
            scanf("%d",&n);
            for(int i=0;i<n;i++)
            {
                scanf("%d",&x);
                ans^=sg[x];
            }
            if(ans) printf("W");
            else    printf("L");
        }
        puts("");
    }
    return 0;
}