Hdu4111Alice and Bob(dp+博弈论)

最新推荐文章于 2021-06-07 17:31:49 发布

原创最新推荐文章于 2021-06-07 17:31:49 发布 · 1k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#博弈论 #dp

数学+DP 专栏收录该内容

116 篇文章

订阅专栏

本文探讨了游戏开发中涉及的策略与算法实现，包括如何使用最佳策略在游戏中取胜，涉及数学逻辑、游戏规则理解及算法优化等内容。

Alice and Bob

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1540 Accepted Submission(s): 563

Problem Description

Alice and Bob are very smart guys and they like to play all kinds of games in their spare time. The most amazing thing is that they always find the best strategy, and that's why they feel bored again and again. They just invented a new game, as they usually did.
The rule of the new game is quite simple. At the beginning of the game, they write down N random positive integers, then they take turns (Alice first) to either:
1. Decrease a number by one.
2. Erase any two numbers and write down their sum.
Whenever a number is decreased to 0, it will be erased automatically. The game ends when all numbers are finally erased, and the one who cannot play in his(her) turn loses the game.
Here's the problem: Who will win the game if both use the best strategy? Find it out quickly, before they get bored of the game again!

Input

The first line contains an integer T(1 <= T <= 4000), indicating the number of test cases.
Each test case contains several lines.
The first line contains an integer N(1 <= N <= 50).
The next line contains N positive integers A ₁ ....A _N(1 <= A _i <= 1000), represents the numbers they write down at the beginning of the game.

Output

For each test case in the input, print one line: "Case #X: Y", where X is the test case number (starting with 1) and Y is either "Alice" or "Bob".

Sample Input

Sample Output

  
   Case #1: Alice
Case #2: Bob
Case #3: Bob

博弈真是博大精深的一门学问啊。后者由前者的状态可以得出结论。后者有多个前者，只要一者满足条件即可判断输赢。详细解题报告见:http://blog.youkuaiyun.com/ahero_happy/article/details/6955595 注意这个相当于打表，越到后面越省时间。状态都存放在sg数组中。(要改掉想题时抓头发的坏习惯。不记得某某某说过了程序猿都是秃子。突然看到草稿本上掉了很多毛发，原来如此。。。)

#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 sg[60][60000];
int dp(int a,int b)
{
	if (sg[a][b]!=-1) 
		return sg[a][b];
	if (b==1) 
		return sg[a][b]=dp(a+1,0);
	sg[a][b]=0;
	if (a>0&&!dp(a-1,b))
		sg[a][b]=1;
	if (b>0&&!dp(a,b-1))
		sg[a][b]=1;
	if (a>0&&b&&!dp(a-1,b+1))
		sg[a][b]=1;
	if (a>1&&((b==0&&!dp(a-2,b+2))||(b&&!dp(a-2,b+3))))
		sg[a][b]=1;
	return sg[a][b];
}
int main()
{
	int T,n,i,x,y,z;
	memset(sg,-1,sizeof(sg));
	scanf("%d",&T);
	for (int c=1;c<=T;c++)
	{	
		y=0;z=0;
		scanf("%d",&n);
		for (i=1;i<=n;i++)
		{
			scanf("%d",&x);
			if (x==1)
				y++;
			else z+=x+1;
		}
		if (z) z--;
		dp(y,z);
		printf("Case #%d: ",c);
		if (sg[y][z]) puts("Alice");
		else puts("Bob");
	}
	return 0;
}