HDU 5724 Chess (博弈 状压+sg函数)

Chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 208    Accepted Submission(s): 67

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 integerspj(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

212 19 2021 191 18

 

Sample Output

NOYES

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

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

题目大意：n行20列的棋盘，对于每行，如果当前棋子右边没棋子，那可以直接放到右边，如果有就跳过放到其后面的第一个空位子，A先操作，最后谁无法操作则输，给定每行棋子状态，问先手是否必胜

题目分析：组合博弈问题，直接sg函数，因为列只有20，可以状压搞，枚举每个状态，找到该状态下可行的操作然后标记，sg函数结论可参考sg函数和sg定理

#include <cstdio>#include <cstring>int const MAX = 21;int sg[1 << MAX], vis[MAX]; int get_sg(int sta){    memset(vis, false, sizeof(vis));    for(int i = 20; i >= 0; i--)    {        if(sta & (1 << i))        {            int tmp = sta;            for(int j = i - 1; j >= 0; j--)            {                if(!(sta & (1 << j)))                {                    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 = 0; i < (1 << 20); i++)        sg[i] = get_sg(i);    int T;    scanf("%d", &T);    while(T --)    {        int n, m, p, ans = 0;        scanf("%d", &n);        for(int i = 0; i < n; i++)        {            scanf("%d", &m);            int sta = 0;            while(m --)            {                scanf("%d", &p);                sta |= (1 << (20 - p));            }            ans ^= sg[sta];        }        printf("%s\n", ans ? "YES" : "NO");    }}