HDU 5724 Chess

Chess

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

Total Submission(s): 674    Accepted Submission(s): 275


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each test case, output one line of “YES” if Alice can win the game, “NO” otherwise.
 

[align=left]Sample Input[/align]

2
1
2 19 20
2
1 19
1 18

 

[align=left]Sample Output[/align]

NO
YES

第一次写的sg、、、

大致印象是由这一状态的所有前驱sg值求出当前状态的sg值，嗯，暂时没有太深的理解了、、、

没有太多了解，也说不出什么。。。留个纪念好了

#include <bits/stdc++.h>
using namespace std;

int sg[1<<22];

void init()
{
for(int i=1;i<(1<<20);i++)
{
int last=-1; //最靠右的0
        int h[25];
memset(h,-1,sizeof(h));
for(int j=0;j<20;j++)
{
if(!((i>>j)&1)) last=j;
else
{
if(last!=-1)//第j位的1可以有last位的1移动而来 last==-1时表示第j为左边没有0
                {
h[sg[i^(1<<j)^(1<<last)]]=1;
}
}
}
int j=0;
while(h[j]!=-1) j++;
sg[i]=j;
}
}

int main()
{
init();
int t,n,m;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int num=0;
for(int i=0;i<n;i++)
{
scanf("%d",&m);
int ans=0;
for(int i=0;i<m;i++)
{
int a;
scanf("%d",&a);
ans^=1<<(20-a);
}
num^=sg[ans];
}
if(num) puts("YES");
else puts("NO");
}
}