uva live 12846 A Daisy Puzzle Game

假设下一个状态有必败。那么此时状态一定是必胜，否则此时状态一定是必败

状压DP

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
int dp[1<<20];
int n;
int dfs(int state)
{
	int i,t;
	if(state==0)//0必败
		return dp[0]=0;
	if(dp[state]!=-1)//此状态已知
		return dp[state];
	dp[state]=0;
	for(i=0;i<n;i++)
	{
		if((state>>i)&1)//摘掉第i个花瓣
		{
			t=state^(1<<i);
			if(dfs(t)==0)
			{
				dp[state]=1;//此状态必胜
				break;
			}
			if(i<n-1&&((t>>(i+1))&1))//再摘掉第i+1个花瓣
			{
				t=t^(1<<(i+1));
				if(dfs(t)==0)
				{
					dp[state]=1;//此状态必胜
					break;
				}
			}
		}
	}
	return dp[state];
}
int main()
{
	bool vis[30];
	int a[30];
	int i,T,j,k,m,t,state,cnt;
	memset(dp,-1,sizeof(dp));
	cin>>T;
	for(j=1;j<=T;j++)
	{
		cin>>n>>m;
		memset(vis,0,sizeof(vis));
		while(m--)
		{
			cin>>t;
			vis[t]=1;
		}
		cnt=0;
		for(i=1;i<=n;i++)
		{
			if(!vis[i])
				a[cnt++]=1;
			else
				break;
		}
		if(i!=n)
		{
			for(k=n;k>=i;k--)
				if(!vis[k])
					a[cnt++]=1;
				else
					a[cnt++]=0;
		}
		state=0;
		for(i=0;i<cnt;i++)
			state+=a[i]*(1<<(cnt-i-1));//生成初始状态
		n=cnt;
		if(dfs(state))
			printf("Case %d: yes\n",j);
		else
			printf("Case %d: no\n",j);
	}
	return 0;
}

12846 A Daisy Puzzle Game

Little Gretchen playing the Daisy game

Gretchen, a little peasant girl from the Swiss Alps, is an expert

at the Daisy game, a simple game that is very well-known

around the country. Two players pluck of the petals of a Daisy

fower, and each player is always at liberty to pluck a single

petal or any two contiguous ones, so that the game would

continue by singles or doubles until the victorious one takes

the last leaf and leaves the “stump”—called the “old maid”—

to the opponent.

The pretty mädchen has mastered the Daisy game to such

an extent that she always plays optimally. In other words, she

always plays by performing the best possible moves on each

turn, a feat which never fails to astonish tourists who dare to

challenge her to a game.

Analyzing the game, it is not very complicated to fgure out a winning strategy for the second player,

as long as the game starts with a complete fower (having all of its petals intact). However, what will

happen when Gretchen plays against an opponent that also plays optimally, and some of the fower’s

petals have been plucked of at random?

A fower is described by a number N which represents the original number of petals of the fower,

and a list of the petals that have been plucked of. All petals are numbered from 1 to N, and given the

circular nature of the fower, that means petals 1 and N are originally adjacent.

Given the description of a fower, and assuming it’s Gretchen’s turn, will she win the game? Remember

that both players always play optimally.

Input

Input starts with a positive integer T, that denotes the number of test cases.

Each test case begins with two integers in a single line, N and M, representing the number of petals

originally in the fower, and the number of petals that have been plucked of, respectively.

The next line contains M distinct integers, representing the petals that have been plucked of. These

numbers will always be in ascending order.

T<= &#20; 5000; 3<= &#20; N<= &#20; 20; 1 &#20;<= M < N

Output

For each test case, print the case number, followed by the string ‘yes’ if Gretchen wins the game, or

‘no’ otherwise.

Sample Input

2

13 1

7

5 3

1 3 4

Sample Output

Case 1: yes

Case 2: no