poj2311(博弈SG函数)

最新推荐文章于 2021-02-22 16:10:29 发布

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本文介绍了一款名为CuttingGame的纸张剪裁博弈游戏，分析了游戏规则及获胜策略。通过水平或垂直剪切纸张，玩家目标是首先剪出单位正方形以赢得比赛。文章提供了基于SG函数的解决方案，并附带源代码。

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Cutting Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4813		Accepted: 1764

Description

Urej loves to play various types of dull games. He usually asks other people to play with him. He says that playing those games can show his extraordinary wit. Recently Urej takes a great interest in a new game, and Erif Nezorf becomes the victim. To get away from suffering playing such a dull game, Erif Nezorf requests your help. The game uses a rectangular paper that consists of W*H grids. Two players cut the paper into two pieces of rectangular sections in turn. In each turn the player can cut either horizontally or vertically, keeping every grids unbroken. After N turns the paper will be broken into N+1 pieces, and in the later turn the players can choose any piece to cut. If one player cuts out a piece of paper with a single grid, he wins the game. If these two people are both quite clear, you should write a problem to tell whether the one who cut first can win or not.

Input

The input contains multiple test cases. Each test case contains only two integers W and H (2 <= W, H <= 200) in one line, which are the width and height of the original paper.

Output

For each test case, only one line should be printed. If the one who cut first can win the game, print "WIN", otherwise, print "LOSE".

Sample Input

2 2
3 2
4 2

Sample Output

LOSE
LOSE
WIN

题意：给一张纸a*b，两人轮流水平或竖直剪，谁先剪到单位正方形谁就赢，求先手的输赢情况。

思路：谁先简称1*x的情况谁就输，所以转换成竖直方向和水平方向两个子问题，再利用SG函数求解。（游戏的和的SG函数值是它的所有子游戏的SG函数值的异或）

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
using  namespace std;
//公平组合博弈SG函数模板 SG[x]相当于Nim博弈中一堆石子的个数
//SG[]:0~n的SG函数值  
//S[]:为x后继状态的集合  
const int MAXN=205;
int SG[MAXN][MAXN],S[MAXN];  
int  getSG(int a,int b)
{
    if(SG[a][b]!=-1)return SG[a][b];
    memset(S,0,sizeof(S));
    for(int i=2;a-i>=i;i++)
    {
    	S[getSG(i,b)^getSG(a-i,b)]=1;
	}
	for(int i=2;b-i>=i;i++)
    {
    	S[getSG(a,i)^getSG(a,b-i)]=1;
	}
	for(int i=0;;i++)
		if(!S[i])return SG[a][b]=i;
} 
int main()
{
	int a,b;
	memset(SG,-1,sizeof(SG));
	while(~scanf("%d%d",&a,&b))
	{
		if(getSG(a,b))printf("WIN\n");
		else printf("LOSE\n");
	}
}