博弈论

最新推荐文章于 2022-10-02 23:40:50 发布

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本文介绍了一款特殊的棋盘游戏，玩家需在一个n行20列的棋盘上进行博弈。通过最优策略判断先手玩家Alice是否能赢得游戏。采用Sprague-Grundy定理计算各状态的SG函数值，最终输出先手玩家能否获胜。

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Chess

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 678 Accepted Submission(s): 277

Problem Description

Alice and Bob are playing a special chess game on an n × 20 chessboard. There are several chesses on the chessboard. They can move one chess in one turn. If there are no other chesses on the right adjacent block of the moved chess, move the chess to its right adjacent block. Otherwise, skip over these chesses and move to the right adjacent block of them. Two chesses can’t be placed at one block and no chess can be placed out of the chessboard. When someone can’t move any chess during his/her turn, he/she will lose the game. Alice always take the first turn. Both Alice and Bob will play the game with the best strategy. Alice wants to know if she can win the game.

Input

Multiple test cases.

The first line contains an integer

T(T≤100), indicates the number of test cases.

For each test case, the first line contains a single integer

n(n≤1000), the number of lines of chessboard.

Then

n lines, the first integer of ith line is

m(m≤20), indicates the number of chesses on the ith line of the chessboard. Then m integers

pj(1≤pj≤20) followed, the position of each chess.

Output

For each test case, output one line of “YES” if Alice can win the game, “NO” otherwise.

Sample Input

Sample Output

NO
YES#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;

const int INF = 0x3f3f3f3f;

int sg[1<<21],s[21];
int n,m,k;


void getsg(int n)
{
    memset(sg,0,sizeof(sg));
    for(int i=1;i<=n;i++)
    {
        memset(s,0,sizeof(s));
        for(int j=20;j>=0;j--)
           {
               if(i&(1<<j))
               {
                 int tmp=i;
                 for(int k=j-1;k>=0;k--)
                 {
                     if(!(i&(1<<k)))
                     {
                         tmp^=((1<<j)^(1<<k));
                         s[sg[tmp]]=1;
                         break;
                     }
                 }
               }
           }
        for(int v=0;v<=20;v++)
        {
            if(s[v]==0)
            {
                sg[i]=v;
                break;
            }
        }
    }
}

int main()
{
    getsg(1<<20);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int ans=0;
        scanf("%d",&n);
        while(n--)
        {
            scanf("%d",&m);
            int sum=0;
            while(m--)
            {
                scanf("%d",&k);
                sum+=(1<<(20-k));
            }
            ans^=sg[sum];
        }
        printf("%s\n", ans ? "YES" : "NO");
    }
    return 0;
}