Codeforces A. Matrix Game (思维) (Round #648 Div.2)

传送门

题意： 有一个棋盘初始状态有些位置是’1’—表示该位置对应的行和列都已经被占用。现在Vivek和Ashish一起做游戏，两人轮流选一个未被占用的位置标记，且Ashish是先手，谁动不了了谁就输了。输出每次的赢家。

思路：

• 直接统计没有’1’的行的数量line，以及没有’1’的列的数量row.
• 取line与row的min为minn，若minn为奇数就是先手赢，反之后手赢。

代码实现：

#include
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n, m, line, row;
int mp[100][100];

signed main()
{
    IOS;

    cin >> t;
    while(t --){
        cin >> n >> m;
        line = 0; row = 0;
        for(int i = 1; i <= n; i ++){
            int ok = 1;
            for(int j = 1; j <= m; j ++){
                cin >> mp[i][j];
                if(mp[i][j]) ok = 0;
            }
            if(ok) line ++;
        }
        for(int j = 1; j <= m; j ++){
            int ok = 1;
            for(int i = 1; i <= n; i ++){
                if(mp[i][j]) ok = 0;
            }
            if(ok) row ++;
        }

        int minn = min(line, row);
//        cout << "line:" << line << " " << "row:" << row << minn << endl;
        if(minn % 2) cout << "Ashish" << endl;
        else cout << "Vivek" << endl;
    }

    return 0;
}