链接:http://acm.hdu.edu.cn/showproblem.php?pid=5724

题意:一个n*20的棋盘,n <= 1000;棋盘上有一些棋子,每颗棋子只能移动到右边的第一个空格。不能移动者输;其中 Alice先手;如果Alice能赢输出"YES";

思路:每个子游戏的大小只有20,使用状压即可;但是每个子游戏的SG值需要建立在比其规模更小的SG值之上;这样,在二进制高位变为0低位变为1之后,显然数值变大了;

那么就二进制低位表示高位编码即可;

预处理出所有的有效状态的SG值之后,使用NIM和看ans 是否为0即可;为0表示先手输;

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define A first
#define B second
#define MK make_pair
#define esp 1e-8
#define zero(x) (((x)>0?(x):-(x))<eps)
#define bitnum(a) __builtin_popcount(a)
#define clear0 (0xFFFFFFFE)
#define mod 1000000007
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
void read1(T &m)
{
T x = 0,f = 1;char ch = getchar();
while(ch <'0' || ch >'9'){ if(ch == '-') f = -1;ch=getchar(); }
while(ch >= '0' && ch <= '9'){ x = x*10 + ch - '0';ch = getchar(); }
m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
if(a>9) out(a/10);
putchar(a%10+'0');
}
inline ll gcd(ll a,ll b){ return b == 0? a: gcd(b,a%b); }
int SG[1<<20], state[22];
void init()
{
for(int i = 0; i < 1<<20; i++){
MS0(state);
rep_1(j,19,0){
if(i & (1<<j)){
rep_1(k,j-1,0) if(!(i & (1 << k))){
state[SG[i^(1<<k)^(1<<j)]] = 1;
break;
}
}
}
rep1(j,0,19) if(state[j] == 0){
SG[i] = j;break;
}
}
}
int main()
{
//freopen("data.txt","r",stdin);
//freopen("out.txt","w",stdout);
init();
int T, kase = 1;
scanf("%d",&T);
while(T--){
int n, m, p, ans = 0;
read1(n);
rep1(i,1,n){
read1(m);
int S = 0;
while(m--){
read1(p);
S |= 1 << 20 - p;
}
ans ^= SG[S];
}
puts(ans?"YES":"NO");
}
return 0;
}

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