HDU1944 S-Nim(博弈论SG函数)
HDU1944 S-Nim
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 rst 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
题解
题意
可以理解为存在m堆的Nim游戏,输出结果,第一行第一个数k,表示有k种取法。
思路
同样还是SG函数的板题,然而多次循环之后会发现会TLE。想到了一个技巧,S中保存的数据如果设置t,每次i++的时候t++,然后每次就不用使用memset清空了,这是一步的优化,对于MAXN较大的情况试用。
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e4+10;
int f[MAXN],SG[MAXN],S[MAXN];
void getSG(int n,int k){
memset(SG,0,sizeof(SG));
int t = 1;
for(int i=1;i<=n;i++){
for(int j=1;j<=k&&f[j]<=i;j++) S[SG[i-f[j]]] = t;
for(int j=0;j<=n;j++){
if(S[j]!=t){
SG[i] = j;
break;
}
}
t++;
}
}
int main(){
int k,m,l;
while(~scanf("%d",&k)){
if(k==0) break;
for(int i=1;i<=k;i++) scanf("%d",&f[i]);
sort(f+1,f+k+1);
getSG(MAXN-5,k);
scanf("%d",&m);
for(int i=0;i<m;i++){
scanf("%d",&l);
int ans = 0;
for(int j=0;j<l;j++){
int t =0;
scanf("%d",&t);
ans ^= SG[t];
}
if (ans==0) printf("L");
else printf("W");
}
printf("\n");
}
return 0;
}