ZOJ 1039 Number Game 记忆化搜索+二进制状态压缩

Number Game Time Limit: 10 Seconds       Memory Limit: 32768 KB

Background

Christiane and Matthias are playing a new game, the Number Game. The rules of the Number Game are:

Christian and Matthias take turns in choosing integer numbers greater than or equal to 2. The following rules restrict the set of numbers which may be chosen:

R1 A number which has already been chosen by one of the players or a multiple of such a number cannot be chosen. (A number z is a multiple of a number y if z can be written as y * x and x is a positive integer.)

R2 A sum of two such multiples cannot be chosen either.

R3 For simplicity, a number which is greater than 20 cannot be chosen either. This enables a lot more NPCs (Non-Personal-Computers) to play this game.

The player who cannot choose any number anymore looses the Number Game.

Here is an example: Matthias starts by choosing 4. Then Christiane is not allowed to choose 4, 8, 12, etc. Let us assume her move is 3. Now, the numbers 3, 6, 9, etc. are excluded, too; furthermore, numbers like: 7 = 3 + 4, 10 = 2 * 3 + 4, 11 = 3 + 2 * 4, 13 = 3 * 3 + 4, . . . are not available. So, in fact, the only numbers left are 2 and 5. Matthias now says 2. Since 5 = 2 + 3 is now forbidden, too, he wins because there is no number for Christiane's move left.

Your task is to write a program which will help to play the Number Game. In general, i.e., without rule R3, this game may go on forever. However, with rule R3, it is possible to write a program that finds a strategy to win the game.

Problem

Given a game situation (a list of numbers which are not yet forbidden), your program should output all winning moves. A winning move is a move by which the player whose turn it is can force a win no matter what the other player will do. Now we define these terms more formally:

A loosing position is a position in which either

1. all numbers are forbidden, or

2. no winning move exists.

A winning position is a position in which a winning move exists.

A winning move is a move after which the position is a loosing position.

Input

The first line contains the number of scenarios.

The input for each scenario describes a game position. It begins with a line containing the number a, 0 <= a < 20 of numbers which are still available. Next follows a single line with the a numbers still available, separated by single blanks.

You may assume that all game positions in the input could really occur in the Number Game (for example, if 3 is not in the list of numbers available, 6 will not be, either).

Output

The output for each scenario begins with a line containing "Scenario #i:" where i is the number of the scenario starting at 1. In the next line either print "There is no winning move." if this is true for the position of the current scenario, or "The winning moves are: w1 w2 . . . wk." where the wi are all the winning moves, in ascending order, separated by single blanks. The output for each scenario should be followed by a blank line.

Sample Input

2
1
2
2
2 3

Sample Output

Scenario #1:
The winning moves are: 2.

Scenario #2:
There is no winning move.

题意：有2到20这个集合序列，告诉一个n个数字的可选序列，两个人从可选序列中选数字，比如说a，那么a的倍数将不能选，a以及a的倍数与已经被标记的数字的和将也不能选，即两个不能取的数字的和也将不能选。只有19个数，通过状态压缩，用dp[1<<19]来保存集合状态即可。对于每一位，1表示可选，0表示不可选。

dp[i]表示对于i这个数表示的状态集合，起始时都初始化为0.。dp[i]=1时，i表示一个状态集合。对于可选的某个数，对他进行遍历，如果选了这个数以后，他所能到达的所有状态都是必输状态，即选了这个数，对手没有数可选了或者选神马都是输，那就把这个数标记为必胜状态。

#include <stdio.h>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

int n;
int a;
int MAX=1<<19;

int win[(1<<19)+10];

int get(int mask,int x)
{
    int ret=mask;
    for(int i=x;i<=20;i+=x)//删除倍数
    {
        ret&=~(1<<(i-2));
    }

    for(int i=2;i<=20;i++)
    {
        if( (ret&(1<<(i-2))))//枚举还存在的数
        {
            for(int j=i-x;j>=2;j-=x)//找是否存在两个非法数和也非法的情况
            {
                if(!(ret&(1<<(j-2))))
                {
                    ret&=~(1<<(i-2));
                    break;
                }
            }
        }
    }

    return ret;
}

int dfs(int mask)
{
    if(win[mask]==0)
    return 0;
    if(win[mask]==1)
    return 1;

    for(int i=2;i<=20;i++)
    {
        if(mask&(1<<(i-2)))
        {
            int tmp=get(mask,i);
            if(!dfs(tmp))//如果后面所有都是必输，那么这一步就是必赢
            {
                win[mask]=1;
                return 1;
            }
        }
    }

    win[mask]=0;
    return 0;
}

int main()
{
    int T;
        int mask=MAX;

     scanf("%d",&T);
    {
        for(int ca=1;ca<=T;ca++)
        {
             memset(win,-1,sizeof win);

    for(int i=0;i<19;i++)
        win[1<<i]=1;

            scanf("%d",&n);

             mask=0;

            for(int i=0;i<n;i++)
            {
                scanf("%d",&a);
                mask|=(1<<(a-2));
            }

           printf("Scenario #%d:\n",ca);

           if(!dfs(mask))//对于整个可选集合进行判断，都是必输，那么就没有必胜局
           {
               printf("There is no winning move.\n\n");
           }
           else
           {
               printf("The winning moves are:");
               for(int i=2;i<=20;i++)
               {
                   if(mask&(1<<(i-2)))
                   {
                       int tmp=get(mask,i);
                       if(!dfs(tmp))//如果之后对方是必输局，则这是必胜选择
                       cout<<" "<<i;
                   }
               }

               puts(".\n");
           }

        }

    }
    return 0;
}