HDU 4642 Fliping game

Fliping game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 123    Accepted Submission(s): 89

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 (x₁, y₁)-(n, m) (1 ≤ x₁≤n, 1≤y₁≤m) and flips all the coins (upward to downward, downward to upward) in it (i.e. flip all positions (x, y) where x₁≤x≤n, y₁≤y≤m)). The only restriction is that the top-left corner (i.e. (x₁, y₁)) 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

2

2 2

1 1

1 1

3 3

0 0 0

0 0 0

0 0 0

 

Sample Output

Alice

Bob

 

Source

2013 Multi-University Training Contest 4

 

 多校第四场1011，一道博弈题问题

由题意很容易发现，每次翻转都会翻动到最右下角的那枚棋子，而获胜的那个人最终把所有棋子都翻到反面去，也必定是翻动了那枚棋子。

因此第一次翻动时最右下角棋子的状态就已经决定了胜负。若为正面则Alice胜，反面则Bob胜

#include<iostream>
#include<cstdio>

using namespace std;

int n,m;
int g[200][200];

int main()
{
    int t;

    cin>>t;

    while(t--)
    {
        scanf("%d %d",&n,&m);

        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&g[i][j]);

        if(g[n][m]==1)
            cout<<"Alice"<<endl;
        else
            cout<<"Bob"<<endl;
    }

    return 0;
}