ACM2016多校联赛1B hdu5724 Chess（博弈）

感谢围巾师兄，不然完全想不到。

Chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1006    Accepted Submission(s): 439


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

思路：每一行有2^20种情况嘛，有或没有，用状态压缩把每一种情况的sg值算出来。

然后把每一行的sg亦或起来就可以了。

上代码加注释吧，不懂的地方请留言，天天在线可以及时回复。

//http://acm.hdu.edu.cn/showproblem.php?pid=5724
//Chess
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=(1<<21);
int sg[maxn];
bool vis[25];
int getsg(int x)
{
memset(vis,false,sizeof(vis));
for(int i=20;i>=0;i--)
{
if(x&(1<<i))//判断该位置是否存在棋子
{
int temp=x;
for(int j=i-1;j>=0;j--)//找到右边第一个为空的位置
{
if(!(x&(1<<j)))
{
vis[sg[temp^=(1<<j)^(1<<i)]]=true;//sg函数公式
break;//一定要break  因为只能最多到达右边第一个为空的位置
}
}
}
}
int ans=0;
while(vis[ans])ans++;
return ans;
}
int main()
{
for(int i=0;i<(1<<20);i++)sg[i]=getsg(i);
int T,n,m,t;
scanf("%d",&T);
while(T--)
{
int res=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&m);
int ans=0;
for(int j=1;j<=m;j++)
{
scanf("%d",&t);
ans|=(1<<20-t);
}
res=res^sg[ans];
}
if(res)cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
}