hdu 5724 Chess 博弈sg+状态压缩

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

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define LL __int64
#define pi (4*atan(1.0))
#define eps 1e-8
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=3e6+,M=1e6+,inf=1e9+;
const LL INF=1e18+,mod=1e9+;

int sg[N];
int dfs(int n)
{
    if(sg[n]!=-)return sg[n];
    int flag[],nex[], mex[];
    for(int i=;i<;i++)
        flag[i]=,mex[i]=,nex[i]=;Q
    for(int i=;i<;i++)
    if(n&(<<i))flag[i]=;
    nex[]=;
    for(int i=;i>=;i--)
    if(!flag[i+])nex[i]=i+;
    else nex[i]=nex[i+];
    for(int i=;i<;i++)
    {
        if(flag[i]&&nex[i]<)
        {
            int z=n-(<<i)+(<<nex[i]);
            //cout<<z<<endl;
            dfs(z);
            mex[sg[z]]=;
        }
    }
    for(int i=;;i++)
        if(!mex[i])return sg[n]=i;
}
int main()
{

    memset(sg,-,sizeof(sg));
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        int ans=;
        for(int i=;i<=n;i++)
        {
            int m,sum=;
            scanf("%d",&m);
            for(int j=;j<=m;j++)
            {
                int x;
                scanf("%d",&x);x--;
                sum+=(<<x);
            }
            ans^=dfs(sum);
        }
        if(ans)printf("YES\n");
        else printf("NO\n");
    }
    return ;
}