题解：B4186 [中山市赛 2024/科大国创杯小学组 2023] 六形棋/海克斯

#include<bits/stdc++.h>
#define N 105
using namespace std;

class Quick_IO
{
  public:
	template<typename T>
	void read(T &r)
	{
		T x = 0, f = 1;
		char ch = getchar();
		while (ch < '0' || ch > '9')
		{
			if (ch == '-') f = -1;
			ch = getchar();
		}
		while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
		r = x * f;
	}
	template<typename T, typename... Args>
	void read(T &tmp, Args &... tmps)
	{
		read(tmp);
		read(tmps...);
	}
} io;

const int mx[7] = {0, 0, 0, 1, 1, -1, -1};
const int my[7] = {0, 1, -1, 0, -1, 0, 1};
int G[N][N];
int T, n;
struct node
{
	int x, y;
};

bool INRANGE(int x, int y)
{
	return (x >= 1 && y >= 1 && x <= n && y <= n);
}

bool BFS(int sx, int sy, bool is_Jimmy)	//起始位置，是否是吉米
{
	bool vis[N][N] = {};	//由于范围不大，直接写临时变量，省的清空
	queue<node> Q;
	Q.push({sx, sy});
	vis[sx][sy] = 1;	//标记走过

	while (Q.size())	//BFS
	{
		int ux = Q.front().x, uy = Q.front().y;
		Q.pop();

		if (is_Jimmy && ux == n) return 1;	//检查是否到终点
		if (!is_Jimmy && uy == n) return 1;

		for (int i = 1, vx, vy; i <= 6; i++)
		{
			vx = ux + mx[i], vy = uy + my[i];
			if (!vis[vx][vy] && G[vx][vy] == (is_Jimmy ? 1 : -1) && INRANGE(vx, vy))
			{
				vis[vx][vy] = 1;
				Q.push({vx, vy});
			}
		}
	}
	return 0;
}

string Solve()	//每一轮的处理
{
	for (int i = 1; i <= n; i++)
	{
		if (G[1][i] == 1) if (BFS(1, i, 1)) return "Jimmy";
		if (G[i][1] == -1) if (BFS(i, 1, 0)) return "Chen";
	}
	return "yet";
}

signed main()
{
	io.read(T);
	while (T--)
	{
		memset(G, 0, sizeof G);	//多测清空
		io.read(n);
		for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) io.read(G[i][j]);
		cout << Solve() << "\n";
	}
}