SG函数(dfs)模板   (测试成功)

代码如下:
int sg[10001];
int f[101];
// f[]为可以取的石子个数,k为f[]的长度
sort(f,f+K);  // f[]必须是从小到大排列
memset(sg,-1,sizeof(sg));  // sg[0]=0, 最小值为0 ,初始化
int Get_SG(int x,int k)  // 计算sg[x]
{
        int visited[101]={0};
        if(sg[x] != -1) return sg[x]; //  同一个集合, 如果已经计算过的sg[x], 可以重复利用
        if(x<f[0]) return sg[x]=0;
        for(int j=0; f[j]<=x && j<k; j++)  // visited 最多更改k次
            visited[Get_SG(x-f[j],k)]=1;
        for(int i=0;;i++)
            if(!visited[i]) return sg[x]=i;
}

 

 

 

题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1536

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3845    Accepted Submission(s): 1685


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
分析: 此题 用到 SG函数 (dfs)模板
#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#define Inf 0x7fffffff // 0x 是数字0,而不是字母o
#define N 10001
using namespace std;
int sg[N],K;
int f[101];
int Get_SG(int x)         
{
        int visited[101]={0};
        if(sg[x] != -1) return sg[x]; //  同一个集合, 如果已经计算过的sg[x], 可以重复利用
        if(x<f[0]) return sg[x]=0;
        for(int j=0; f[j]<=x && j<K; j++)
            visited[Get_SG(x-f[j])]=1;
        for(int i=0;;i++)
            if(!visited[i]) return sg[x]=i;
}
int main(){
   int l,temp,x,m,Max;
   while(scanf("%d",&K) && K)
   {
       int x=0,Max=0;
       for(int i=0;i<K;i++)
           scanf("%d",&f[i]);
        sort(f,f+K);
        scanf("%d",&m);
        memset(sg,-1,sizeof(sg));
       while(m--)
       {
           scanf("%d",&l);
           x=0;
           for(int i=0 ;i<l; i++)
           {
               scanf("%d",&temp);
               x^=Get_SG(temp);
           }
           if(x) printf("W");
           else printf("L");
       }
       printf("\n");
   }
    return 0;
}