hdu 5724 Chess(博弈)

Chess



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(T100), indicates the number of test cases.

For each test case, the first line contains a single integer n(n1000), the number of lines of chessboard.

Then n lines, the first integer of ith line is m(m20), indicates the number of chesses on the ith line of the chessboard. Then m integers pj(1pj20)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
 
 
将棋盘的状态压缩,然后预处理出所有状态的sg值,然后就是普通的组合游戏做法了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int sg[(1<<20) + 1010];
int vis[1010];
void init(int x) {
    sg[1] = 0;
    memset(vis, 0, sizeof(vis));
    for(int i = 20; i >= 0; i--) {
        if((1 << i) & x) {
            for(int j = i - 1; j >= 0; j--) {
                if(!((1 << j) & x)) {
                    vis[sg[x ^ (1 << j) ^ (1 << i)]] = 1;
                    break;
                }
            }
        }
        for(int j = 0; ; j++) if(!vis[j]) {
            sg[x] = j;
            break;
        }
    }
}
int main() {
    int t;
    for(int i = 0; i < (1 << 20); i++) init(i);
    scanf("%d", &t);
    while(t--) {
        int n, m;
        scanf("%d", &n);
        int ans = 0;
        for(int i = 1; i <= n; i++) {
            scanf("%d", &m);
            int tmp = 0;
            for(int j = 1; j <= m; j++) {
                int x;
                scanf("%d", &x);
                tmp ^= 1 << (20 - x);
            }
            ans ^= sg[tmp];
        }
        if(ans) puts("YES");
        else puts("NO");
    }
}

 

posted @ 2016-07-21 09:44 MartinEden 阅读(...) 评论(...) 编辑 收藏