【博弈+找规律】HDU_4642_Fliping game

原创于 2017-08-05 23:07:05 发布 · 252 阅读

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本文介绍了一种名为翻转游戏的特殊棋盘游戏，玩家Alice和Bob轮流选择棋盘上的矩形区域进行硬币翻转操作，目标是在遵循特定规则的情况下使所有硬币朝下。文章探讨了最优策略下哪位玩家将赢得比赛。

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Fliping game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2029 Accepted Submission(s): 1363

Problem Description

Alice and Bob are playing a kind of special game on an N*M board (N rows, M columns). At the beginning, there are N*M coins in this board with one in each grid and every coin may be upward or downward freely. Then they take turns to choose a rectangle (x ₁, y ₁)-(n, m) (1 ≤ x ₁≤n, 1≤y ₁≤m) and flips all the coins (upward to downward, downward to upward) in it (i.e. flip all positions (x, y) where x ₁≤x≤n, y ₁≤y≤m)). The only restriction is that the top-left corner (i.e. (x ₁, y ₁)) must be changing from upward to downward. The game ends when all coins are downward, and the one who cannot play in his (her) turns loses the game. Here's the problem: Who will win the game if both use the best strategy? You can assume that Alice always goes first.

Input

The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case starts with two integers N and M indicate the size of the board. Then goes N line, each line with M integers shows the state of each coin, 1<=N,M<=100. 0 means that this coin is downward in the initial, 1 means that this coin is upward in the initial.

Output

For each case, output the winner’s name, either Alice or Bob.

Sample Input

Sample Output

      
       Alice
Bob

Source

2013 Multi-University Training Contest 4

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#include <bits/stdc++.h>

using namespace std;
int a[110][110];
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        if(a[n][m]==1)
            printf("Alice\n");
        else printf("Bob\n");
    }
    return 0;
}