HDU1851(sg博弈+nim博弈)

最新推荐文章于 2023-07-11 00:59:15 发布

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#博弈论

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本文介绍了一种基于博弈论的取石子游戏算法实现，玩家轮流从不同堆中取石子，取最后一颗石子者获胜。文章通过样例输入输出展示了如何使用SG函数确定最优策略。

A Simple Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1107 Accepted Submission(s): 689

Problem Description

Agrael likes play a simple game with his friend Animal during the classes. In this Game there are n piles of stones numbered from 1 to n, the 1st pile has M₁ stones, the 2nd pile has M₂ stones, ... and the n-th pile contain M_n stones. Agrael and Animal take turns to move and in each move each of the players can take at most L₁stones from the 1st pile or take at most L₂ stones from the 2nd pile or ... or take L_n stones from the n-th pile. The player who takes the last stone wins.

After Agrael and Animal have played the game for months, the teacher finally got angry and decided to punish them. But when he knows the rule of the game, he is so interested in this game that he asks Agrael to play the game with him and if Agrael wins, he won't be punished, can Agrael win the game if the teacher and Agrael both take the best move in their turn?

The teacher always moves first(-_-), and in each turn a player must takes at least 1 stones and they can't take stones from more than one piles.

Input

The first line contains the number of test cases. Each test cases begin with the number n (n ≤ 10), represent there are n piles. Then there are n lines follows, the i-th line contains two numbers M_i and L_i (20 ≥ M_i > 0, 20 ≥ L_i > 0).

Output

Your program output one line per case, if Agrael can win the game print "Yes", else print "No".

Sample Input

Sample Output

Yes
No

题意大概是一个老师跟一个学生过不去，要通过取石子决定成不惩罚他们。老师先取，有n堆石子，第i堆有Mi个，第i堆最多只能取Li个。最后取完者胜。

博弈论写得越来越熟练了。将每堆的sg值求出来，总异或操作即可。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
using namespace std;
#define LL __int64
int m,l,T,n;
int sg[25];
int mex(int x)
{
	int i;
	if (sg[x]!=-1) return sg[x];
	bool vis[22];
	memset(vis,0,sizeof(vis));
	for (i=1;i<=l;i++)
	{
		int t=x-i;
		if (t<0) break;
		sg[t]=mex(t);
		vis[sg[t]]=1;
	}
	for (i=0;;i++)
	if (!vis[i])
	{
		sg[x]=i;
		break;
	}
	return sg[x];
}
int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n);
		memset(sg,0,sizeof(sg));
		int ans=0;
		for (int i=1;i<=n;i++)
		{
			memset(sg,-1,sizeof(sg));
			scanf("%d%d",&m,&l);
			ans^=mex(m);
		}
		if (ans) puts("No");
		else puts("Yes");
	}
	return 0;
}