感觉这题非常牛逼，写个题解纪念一下。其实就是抄写 Itst 博客。 显然黑色不可能赢。 先假设没有提前涂白的点。 考虑前 \(O(1)\) 步白色必胜的情况： 存在点的度数 \(\geq 4\)。 存在点的度数 \(=3\)，并且所连的 \(3\) 个点中至少有 \(2\) 个非叶子节点。 其余情况，树的形态就只
class TicTacToe { int n; int[] rows; int[] cols; int diag=0; int antiDiag = 0; public TicTacToe(int n) { this.n = n; rows = new int[n]; cols = new int[n]; } public int move(int row, int col, in
Assume the following rules are for the tic-tac-toe game on an n x n board between two players: A move is guaranteed to be valid and is placed on an empty block. Once a winning condition is reached, no more moves are allowed. A player who succeeds in plac
题目链接 题目 Certainly, everyone is familiar with tic-tac-toe game. The rules are very simple indeed. Two players take turns marking the cells in a 3 × 3 grid (one player always draws crosses, the other — noughts). The player who succeeds first in placin
#include <cstdio> #include <cstring> const int MAXN = 300; int T , n , k , cnt; char str[ MAXN + 5 ][ MAXN + 5 ] , Ans[ MAXN + 5 ][ MAXN + 5 ]; int main( ) { scanf("%d",&T); while( T -- ) { scanf("%d",&n); k =
https://www.cnblogs.com/zafu/p/10557800.html 2016年09月01日 鉴于研究所的需求，最近开始研究Chelsio T5（终结者5），本篇博文对相关技术的基础概念做了罗列，并给了一些扩展学习链接。后续自己将针对RDMA技术做进一步学习和研究！ 核心基础概念 FCoE：以太网光纤通道 (Fibre Chan
这题在 CF rating 是 3100+，听了讲评之后感觉醍醐灌顶。 如果您不看题解就 AC，那您是真的强。 首先，我们发现，黑不可能赢。 接下来，考虑一种简单的情况：没有任何点初始时有颜色。 情况 1：树中有一个点 \(A\) 的度大于等于 \(4\)。 我们假设它连着 \(B,C,D,E\) 等点。那么，白方下 \(A\)，不妨
原题链接https://codeforces.com/problemset/problem/3/C 题目本身不难，但是细节很多 题意：给你一个井字棋盘，要求你判断其状态 思路：按照规则判断即可，难的是不合法状态考虑不全，先手和后手赢的时候，两个人棋盘上的棋子关系是固定的。 代码如下 char g[5][5]; bool flag = false; //判断
Description The only difference between the easy and hard versions is that tokens of type O do not appear in the input of the easy version. Errichto gave Monogon the following challenge in order to intimidate him from taking his top contributor spot on Co
注意在循环中在try语句中使用nextInt()方法至少两次以上时，如果在第一次nextInt() catch exception 并重新执行循环，那么接下来的nextInt（）执行的是之前输入行未读取晚剩下的部分， 有可能进入循环。解决方法是 先全部读取成，然后再判断. private void nextCell(char[][] cells, Scanne
package LeetCode_348 /** * 348. Design Tic-Tac-Toe * (Lock by leetcode) * https://www.lintcode.com/problem/design-tic-tac-toe/description * Design a Tic-tac-toe game that is played between two players on a n x n grid. You may assume the following ru
能源必知的油气当量，标准煤换算 http://www.360doc.com/content/19/0513/18/63592782_835459328.shtml1000-1255立方米天然气等于一吨当量石油一吨石油 等于 7.3桶石油一吨石油 等于1.5吨原煤大概的比率。 油气当量就是将天然气产量按热值折算
原题链接在这里：https://leetcode.com/problems/design-tic-tac-toe/ 题目： Design a Tic-tac-toe game that is played between two players on a n x n grid. You may assume the following rules: A move is guaranteed to be valid and is placed on an empty block. Once
【收藏】 能源必知的油气当量，标准煤换算 https://www.sohu.com/a/219096462_738536 油气当量就是将天然气产量按热值折算为原油产量的换算系数。标准油气当量，是根据原油和天然气的热值折算而成的油气产量，一般取1255立方米天然气=1吨原油，通常为了简化，取1000立方米天然
Tic Tac Toe is a child’s game played on a 3 by 3 grid. One player, X, starts by placing an X at an unoccupied grid position. Then the other player, O, places an O at an unoccupied grid position. Play alternates between X and O until the grid is filled
Source : https://www.hotrod.com/articles/ctrp-0407-ackermann-steering-system/ Tuning Your Steering System To Reduce Drag One of the critical elements of proper chassis alignment is the steering system. The front wheels must work together, just like the
https://ac.nowcoder.com/acm/contest/847/B #include<algorithm>#include<cstring>#include<iostream>#include<math.h>#include<string>#include<stdio.h>#include<map>#include<queue>#define ll long long#define inf 0
题面: 翻译: 有一些牛在玩一个游戏: 他们把一些字母写在一个3*3的黑板上. 这些奶牛的名字的首字母分别是A…Z 如果黑板上有相同的字母能够连成一行/一列/一斜.那么名字首字母为这个字母的牛获胜. 但是这样的话太难获胜了,所以这个游戏可以允许两个人合作. 如果有两个字母可以
