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.
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(" "); } return 0; }