hdu 5724 SG+状态压缩

Chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1821    Accepted Submission(s): 799

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

/*
hdu 5724 SG+状态压缩

感觉上是博弈，而且很久以前就看了看SG，但是并没怎么系统地去学习zzz。
首先可以把棋盘n行看成n个石碓，用1表示有棋子，0表示没有的话，能够用二进制表示出所有的状态：
1000100这个可以转换成 0100100 1000010等等

然后就能利用公式求出每种情况的SG(枚举 1~(1<<20))
得出每一行的状态计算即可

hhh-2016-08-01 17:28:17
学习：
//http://blog.csdn.net/luomingjun12315/article/details/45555495
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <vector>
#include <queue>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
using namespace std;
const ll mod = 1e9 + 7;
const ll INF = 0x3f3f3f3f;
const int maxn = 1000100;

int vis[22];
int sg[1<<20];

int SG(int cur)
{
    memset(vis,0,sizeof(vis));
    for(int i = 20; i >= 0; i--)
    {
        if(cur & (1<<i))
        {
            for(int j = i-1; j >= 0; j--)
            {
                if(!(cur & (1 << j)))
                {
                    int tmp = cur;
                    tmp ^= ((1<<i)^(1<<j));
                    vis[sg[tmp]] = true;
                    break;
                }
            }
        }
    }
    for(int i = 0 ; i <= 20; i++)
    {
        if(!vis[i])
            return i;
    }
    return 0;
}

int main()
{
    memset(sg,0,sizeof(sg));
    for(int i = 1; i < (1 << 20); i++)
        sg[i] = SG(i);
    int T,n,x;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int ans = 0;
        for(int i = 1;i <= n;i++)
        {
            int m,cur = 0;
            scanf("%d",&m);
            for(int j = 1;j <= m;j++)
            {
                scanf("%d",&x);
                cur |= 1 << (20-x);
            }
            ans ^= sg[cur];
        }
        if(ans )
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}