题目：
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

2016多校第一场的博弈题，作为一个不会状压的鶸今天看到题解赶紧滚去学习了下，果然博大精深，比赛时的求sg值的思路有问题所以GG了，结合网上大神思路改了下：

    利用状压的思想求出当前状态的所有后继状态并根据他们的sg值确定当前状态的sg值；

    因为移动后表示状态的数必然变小所以正向遍历；

AC代码…：

#include <bits/stdc++.h>
#define rep(i,j,k) for (int i = j; i <= k; i++)
using namespace std;
int sg[(1<<20)+100];

int get_sg (int i) {
    int h[25];
    memset(h, 0, sizeof(h));
    int last = -1;
    rep(j, 0, 19) {
        if(!((i>>j)&1)) last = j; //依次找到一个0位置，标记
        else { //找到1位置
            if(last!=-1){
                h[sg[i^(1<<j)^(1<<last)]] = 1; //标记后继状态（将标记的0和1互换后的状态）
            }
        }
    }
    int j = 0;
    while(h[j])
        j++;
    return j;  //统计获得sg值
}
int main(){
    int t, n, m, x;
    rep(i, 1, (1<<20))
        sg[i] = get_sg(i);
    scanf("%d",&t);
    while(t--){
        scanf("%d", &n);
        int ans = 0;
        rep(i, 1, n) {
            scanf("%d", &m);
            int t=0;
            rep(i, 1, m){
                scanf("%d", &x);
                t ^= 1<<(20-x); //计算当前的状态
            }
            ans ^= sg[t];
        }
        puts(ans ? "YES" : "NO");
    }
    return 0;
}

太TM菜了，，滚去做题