Fliping game HDU

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

大致题意：就是给你一个01矩阵，然后两个人轮流选取一个数值为1的点，然后将每次选择的点到右下角的矩阵01翻转，谁先无法操作谁输。Bob先手

思路：可以很容易的看出，不论选取的是哪个点，都会对最右下角的那个点造成影响。所以取胜的策略就是使最右下角的那个值变成0后留给对方。这样对方不论怎么选，都会改变右下角的那个值使得变成1留给你。如此反复，直到对方无法选择。所以我们只需判断下右下角的初始值即可。

代码如下

#include <iostream> #include <cmath>#include <algorithm>#include <cstring>#include <queue>#include <cstdio>using namespace std; #define ll long long int int main() {      int t;    cin>>t;    while(t--)    {        int n,m;        int x;        cin>>n>>m;        for(int i=1;i<=n;i++)        for(int j=1;j<=m;j++)        cin>>x;        if(x)        cout<<"Alice\n";        else         cout<<"Bob\n";    }    return 0;}