题解 P6369 【[COCI2006-2007#6] MARATON】

这题我用的是string

首先横着竖着都好弄!重点在于斜的

于是我想到了个方法，只要找到一个不为"."，就判断是否有斜线三个，20ms，代码：

#include <iostream>
#include <cstring>
using namespace std;

string s[35];

int main()
{
    int n;
    cin >> n;
    for(int i = 1; i <= n; i++)
    {
        cin >> s[i];
    }
    //横斜一起，纵单独
    for(register int i = 1; i <= n; i++)
    {
        for(register int j = 0; j <= n - 1; j++)
        {
            if(s[i][j] != '.')
            {
                if(s[i][j] == s[i][j + 1] && s[i][j] == s[i][j + 2])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
                else if(s[i][j] == s[i + 1][j - 1] && s[i][j] == s[i + 2][j - 2])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
                else if(s[i][j] == s[i - 1][j + 1] && s[i][j] == s[i - 2][j + 2])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
                else if(s[i][j] == s[i - 1][j + 1] && s[i][j] == s[i + 1][j - 1])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
                else if(s[i][j] == s[i - 1][j - 1] && s[i][j] == s[i - 2][j - 2])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
               else if(s[i][j] == s[i + 1][j + 1] && s[i][j] == s[i + 2][j + 2])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
                else if(s[i][j] == s[i - 1][j - 1] && s[i][j] == s[i + 1][j + 1])
                {
                    cout << s[i][j] << endl;
                    return 0;
                }
            }

        }
    }
    //纵
    for(register int i = 0; i <= n - 1; i++)
    {
        for(register int j = 1; j <= n; j++)
        {
            if(s[j][i] == s[j + 1][i] && s[j][i] == s[j + 2][i])
            {
                if(s[j][i] != '.')
                {
                    cout << s[j][i] << endl;
                    return 0;
                }
            }
        }
    }
    cout << "ongoing\n";
    return 0;
}

那么if为什么这么多呢？

想想看，如果s[i][j]是三斜线的一个点，那情况是很多的哦，一定要全写到，不然会WA的，况且测试点可有10个！