HDU4023 Game The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

最新推荐文章于 2015-06-02 17:00:26 发布

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本文介绍了一个基于特定Tetris-like板面的游戏策略分析问题。两个玩家轮流放置不同形状的方块，目标是最大化自己占据的空间并阻止对手。文章详细描述了游戏规则、输入输出格式及示例，并提供了一段C++代码实现模拟过程来决定最优策略下哪位玩家将获胜。

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 9 Accepted Submission(s): 4

Problem Description

Alice and Bob are playing game with each other. They play the game on a 2D board. Alice has many vertical 1*2 tiles while Bob has many horizontal 2*1 tiles. They take turn to place their own tiles on the board. Considering about that the tiles cannot overlap each other, the player cannot do the placement any more loses. Since this is such a complex game that they could not find optimal method to play that, Alice decide to simplify this game by replace the large 2D board by some small ones. Alice set up a lot of Tetris tiles instead of the original 2D board. In the other words, the player can only place their own vertical or horizontal tiles on the Tetris-like board. Each player can choose one possible place on any Tetris tiles to place its own tiles. In fact, there are following 15 types of Tetris playground.

The playground cannot be transformed in any ways, including reflection and rotation.
Given the number of each type of tiles, you are asked to determine who will win the game if Alice plays first and both players are playing optimal.

Input

There are multiple test cases; the first line of input contains a single integer denoting the number of test cases.
For each test case, there are only one line contains 15 integers denoting the number of Tetris tiles of the above 15 types. All the numbers are no greater than 100.

Output

For each test cases, output “Alice” if Alice will win the game and both player plays optimally, “Bob” otherwise.

Sample Input

3
5 4 0 0 0 0 0 0 0 0 0 0 0 0 0
5 5 0 0 0 0 0 0 0 0 0 0 0 0 0
100 100 0 0 0 0 0 0 0 0 0 2 1 0 0

Sample Output

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

Source

The 36th ACM/ICPC Asia Regional Shanghai Site —— Online Contest

Recommend

lcy

两个人分别对每种tiles分配一个优先级,

分别计算A选第i中能覆盖几次，同时B在A选i后能覆盖第i种几次，反之亦是如此。

然后模拟计算两人覆盖的次数，比较谁覆盖次数大。

代码：

#include<cstdio>

int p1[]={15,5,6,3,4,11,12,13,14,7,8,9,10,1};
int c1[]={2,0,1,1,2,2,1,1,1,1,1,1,1,1,2};
int d2[]={0,2,0,0,0,0,0,0,1,1,0,0,0,0,0};
int p2[]={15,3,4,5,6,11,12,13,14,9,10,7,8,2};
int c2[]={0,2,2,2,1,1,1,1,1,1,1,1,1,1,2};
int d1[]={2,0,0,0,0,0,1,1,0,0,0,0,0,0,0};
int t,tt,a[25];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		int i,j,cur=0,flag=1,b[2][25]={0},s1=0,s2=0;
		for(i=1;i<=15;i++)
			scanf("%d",&a[i]);
		int f1=0,f2=0;
		while(1)
		{
			int f=1;
			if(cur==0)
			{
				for(i=0;i<14;i++)
					if(a[p1[i]]>0)
					{
						f=0;
						break;
					}
				if(f)
					f1=1;
				if(f1&&f2)
					break;
				a[p1[i]]--;
				s1+=c1[p1[i]-1];
				s2+=d2[p1[i]-1];
			}
			else
			{
				for(i=0;i<14;i++)
					if(a[p2[i]]>0)
					{
						f=0;
						break;
					}
				if(f)
					f2=1;
				if(f1&&f2)
					break;
				a[p2[i]]--;
				s2+=c2[p2[i]-1];
				s1+=d1[p2[i]-1];
			}
			cur^=1;
		}
		printf("Case #%d: %s\n",++tt,s1>s2?"Alice":"Bob");
	}
}