Matrix Game

最新推荐文章于 2020-10-29 22:27:40 发布

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本文探讨了一种基于矩阵的石头堆叠游戏策略，通过分析每一行石头总数并运用妮姆博弈理论，确定了Alice与Bob两人游戏的胜者。文章提供了一个C++实现示例，展示了如何计算游戏结果。

题目

Given an m x n matrix, where m denotes the number of rows and n denotes the number of columns and in each cell a pile of stones is given. For example, let there be a 2 x 3 matrix, and the piles are

2 3 8

5 2 7

That means that in cell(1, 1) there is a pile with 2 stones, in cell(1, 2) there is a pile with 3 stones and so on.

Now Alice and Bob are playing a strange game in this matrix. Alice starts first and they alternate turns. In each turn a player selects a row, and can draw any number of stones from any number of cells in that row. But he/she must draw at least one stone. For example, if Alice chooses the 2nd row in the given matrix, she can pick 2 stones from cell(2, 1), 0 stones from cell (2, 2), 7 stones from cell(2, 3). Or she can pick 5 stones from cell(2, 1), 1 stone from cell(2, 2), 4 stones from cell(2, 3). There are many other ways but she must pick at least one stone from all piles. The player who can't take any stones loses.

Now if both play optimally who will win?

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers: m and n (1 ≤ m, n ≤ 50). Each of the next m lines contains n space separated integers that form the matrix. All the integers will be between 0 and 109 (inclusive).

Output

For each case, print the case number and 'Alice' if Alice wins, or 'Bob' otherwise.

样例

Sample Input

2 3

2 3 8

5 2 7

2 3

1 2 3

3 2 1

Sample Output

Case 1: Alice

Case 2: Bob

妮姆博弈的板子题，只需要把每行的数字先加起来就好了！

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <string.h>
#include <string>
typedef long long ll;
#define INF 0x3f3f3f3f
using namespace std;
int x[55];
int main() {
	int t, casenum = 0;
	int m, n;
	ll sum;
	ll ans;
	cin >> t;
	while (t--){
		cin >> m >> n;
		ans = sum = 0;
		while (m--) {
			sum = 0;
			for (int j = 0; j < n; j++){
				cin >> x[j];
				sum += x[j];
			}
			ans ^= sum;
		}
		if (ans) cout << "Case " << ++casenum << ": Alice" << endl;
		else cout << "Case " << ++casenum << ": Bob" << endl;
	}
	return 0;
}