hdu5724——chess（博弈）（状态压缩dp+sg函数）

最新推荐文章于 2024-08-22 20:10:34 发布

mosquito_zm

最新推荐文章于 2024-08-22 20:10:34 发布

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分类专栏：博弈文章标签：博弈 acm dp

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本文介绍了一个特殊的棋盘游戏，通过组合博弈理论中的sg函数和状态压缩动态规划解决该问题。游戏中，玩家在一个n行20列的棋盘上移动棋子，目标是在对手无法移动时获胜。文章提供了详细的算法实现过程。

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.

大意：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。

题解：组合博弈问题，应用sg函数+状态压缩dp可解。
sg函数可看白书的组合游戏部分。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = (1 << 20) + 10;
int SG[maxn];
int cnt = 0;
int n,m;   
int dfs(int k)
{
    if(SG[k]!=-1) return SG[k];
    int vis[30];
    memset(vis,0,sizeof(vis));
    for(int i=0;i<19;i++){
        int temp = k;
        if(k & (1 << i) == 0) continue;
        if((k & (1 << i))) {
            for(int j=i+1;j<=19;j++){
                if((k & (1 << j)) == 0) {
                    temp = k;
                    temp ^= (1 << i);
                    temp ^= (1 << j);
                    vis[dfs(temp)] = 1; 
                    break;
                }
            }
        }
    }
    for(int i = 0;i < 29;i++) 
      if(vis[i]==0) {
        SG[k] = i;
        return i;
      }
}
int main()
{
    freopen("B.in","r",stdin);
    freopen("B.out","w",stdout); 
    memset(SG,-1,sizeof(SG));
    int temp = 0;
    for(int i=19;i>=0;i--){
        temp += (1 << i);
        SG[temp] = 0;
    }
    int kase;scanf("%d",&kase);
    while(kase--){
        int temp = 0,a,ans = 0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            temp = 0;
            scanf("%d",&m);
            for(int j=1;j<=m;j++){
                scanf("%d",&a);
                temp ^= (1 << (a-1));
            }
            int flag = dfs(temp);
            ans ^= SG[temp];
        }
        if(ans != 0) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}