CodeForces_1450C Errich-Tac-Toe(构造)_codeforces 1450-优快云博客

本文介绍了一种Tic-Tac-Toe变种游戏的解决方案，目标是在限定的操作次数内将游戏状态转变为平局。通过巧妙地利用网格染色策略，确保任意一行或一列不会出现连续三个相同标记，从而达成游戏要求。

Errich-Tac-Toe

time limit per test：1 seconds memory limit per test：256 megabytes

Problem Description

Errichto gave Monogon the following challenge in order to intimidate him from taking his top contributor spot on Codeforces.

In a Tic-Tac-Toe grid, there are $n$ rows and $n$ columns. Each cell of the grid is either empty or contains a token. There are two types of tokens: X and O. If there exist three tokens of the same type consecutive in a row or column, it is a winning configuration. Otherwise, it is a draw configuration.
在这里插入图片描述
In an operation, you can change an X to an O, or an O to an X. Let $k$ denote the total number of tokens in the grid. Your task is to make the grid a draw in at most $⌊k3⌋\left\lfloor\frac{k}{3}\right\rfloor$ (rounding down) operations.

You are not required to minimize the number of operations.

Input

The first line contains a single integer $t$ ( $1 \leq t \leq 100$ ) — the number of test cases.

The first line of each test case contains a single integer $n$ ( $1 \leq n \leq 300$ ) — the size of the grid.

The following $n$ lines each contain a string of $n$ characters, denoting the initial grid. The character in the $i$ -th row and $j$ -th column is ‘.’ if the cell is empty, or it is the type of token in the cell: ‘X’ or ‘O’.

It is guaranteed that not all cells are empty.

The sum of $n$ across all test cases does not exceed 300.

Output

For each test case, print the state of the grid after applying the operations.

We have proof that a solution always exists. If there are multiple solutions, print any.

Sample Input

3
3
.O.
OOO
.O.
6
XXXOOO
XXXOOO
XX…OO
OO…XX
OOOXXX
OOOXXX
5
.OOO.
OXXXO
OXXXO
OXXXO
.OOO.

Sample Output

.O.
OXO
.O.
OXXOOX
XOXOXO
XX…OO
OO…XX
OXOXOX
XOOXXO
.OXO.
OOXXO
XXOXX
OXXOO
.OXO.

题意

在 $n * n$ 的网格图中，部分网格被填充上了’X’令牌或’O’令牌。在一行或一列中，如果被填充上了连续三个相同类型的令牌，则称当前局面为胜局，否则称为平局。现在给出一种游戏的局面，网格中共有 $k$ 个令牌，最多进行 $⌊k3⌋\left\lfloor\frac{k}{3}\right\rfloor$ 次操作，将其变为平局。给出一种合理解。

思路

设下标为 $(i, j)$ ，对网格进行染色，根据 $(i+j)%3(i+j)\%3$ 讲网格染成0,1,2三种不同的颜色。染色后，若某行或某列有三个连续的元素，则三个元素必然属于不同的颜色。所以考虑让某种颜色不包含‘O’，某种颜色不包含’X’即可。
C1：根据网格颜色，将元素放入vector中，求出元素最少的vector，将其中所有元素变为‘O’即可。
设三种颜色的数量分别为num0，num1，num2，且num0最小，则 $3∗num0≤num0+num1+num2=k3*num0\le num0+num1+num2=k$

C2: 根据网格颜色，令牌符号将元素放入vector中，将某种颜色的网格中所有的’X’变成‘O’，某种颜色的网格中所有的’O’变成‘X’,这样则满足一种颜色的网格中没有’X’,一种颜色的网格中没有’O’。求出最小的 $id_1,id_2(id_1!=id_2)$ ，满足 $g[id_1][0].size()+g[id_2][1].size()$ 最小。
设vector大小分别为num00,num01,num10,num11,num20,num21,假设 $id_1 = 0, id_2 = 1$ ，则 $3∗(num00+num11)≤(num00+num11)+(num10+num21)+(num20+num01)=k3*(num00+num11)\le (num00+num11)+(num10+num21)+(num20+num01) = k$ 。

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<iterator>
#define dbg(x) cout<<#x<<" = "<<x<<endl;
#define INF 0x3f3f3f3f
#define LLINF 0x3f3f3f3f3f3f3f3f
#define eps 1e-6
 
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 320;
const int mod = 998244353;
char str[maxn][maxn];
vector<P> g[4][4];

int main()
{
    int t, n, m, i, j, k;
    scanf("%d", &t);
    while(t--)
    {
        for(i=0;i<4;i++){
            g[i][0].clear();
            g[i][1].clear();
        }
        scanf("%d", &n);
        for(i=0;i<n;i++)
            scanf("%s", str[i]);
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                if(str[i][j] == 'O')g[(i+j)%3][0].push_back(P(i,j));
                else if(str[i][j] == 'X') g[(i+j)%3][1].push_back(P(i, j));
        int id1, id2, mx = INF;
        for(i=0;i<3;i++)
            for(j=0;j<3;j++)
                if(i!=j && g[i][0].size()+g[j][1].size()<mx){
                    mx = g[i][0].size()+g[j][1].size();
                    id1 = i, id2 = j;
                }
        for(i=0;i<g[id1][0].size();i++){
            P p = g[id1][0][i];
            str[p.first][p.second] = 'X';
        }
        for(i=0;i<g[id2][1].size();i++){
            P p = g[id2][1][i];
            str[p.first][p.second] = 'O';
        }
        for(i=0;i<n;i++)
            printf("%s\n", str[i]);
    }
    return 0;
}