题目链接：

Chess

Time Limit: 2000/1000 MS (Java/Others)   

 Memory Limit: 65536/65536 K (Java/Others)

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

2

1

2 19 20

2

1 19

1 18

Sample Output

NO

YES

题意：

给一个n*20的棋盘,每行的一部分格子里面有棋子,现在每次可以选一个棋子把它挪到右边第一个空位上;现在问先手是否能必胜;

思路：

组合博弈的内容,先处理出所有状态的sg函数值,然后取异或值判断是否为零;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
//#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=2e6+10;
const int maxn=500+10;
const double eps=1e-8;

int sg[N],vis[25];

inline int get_sg(int x)
{
    mst(vis,0);
    //cout<<x<<endl;
    for(int i=19;i>=0;i--)
    {
        if(x&(1<<i))
        {
            int temp=x;
            for(int j=i-1;j>=0;j--)
            {
                if(!(x&(1<<j)))
                {
                    temp^=(1<<i)^(1<<j);
                    //cout<<temp<<endl;
                    vis[sg[temp]]=1;
                    break;
                }
            }
        }
    }
    for(int i=0;i<=19;i++)if(!vis[i])return i;
    return 0;
}
inline void Init()
{
    For(i,0,(1<<20)-1)sg[i]=get_sg(i);
}
int main()
{
        Init();
        int t;
        read(t);
        while(t--)
        {
            int n,a,m,ans=0;
            read(n);
            while(n--)
            {
                int sum=0;
                read(m);
                For(i,1,m)
                {
                    read(a);
                    sum|=(1<<(20-a));
                }
                ans^=sg[sum];
            }
            if(ans)printf("YES\n");
            else printf("NO\n");
        }

        return 0;
}