本题链接：201512-2 消除类游戏

本博客给出本题截图：

C++

#include <iostream>

using namespace std;

const int N = 35;

int n, m;
int g[N][N];
bool st[N][N];

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < m; j ++ )
            cin >> g[i][j];

    for (int i = 0; i < n; i ++ )
        for (int j = 0; j < m; j ++ )
        {
            int l = j, r = j, u = i, d = i, x = g[i][j];
            while (l >= 0 && g[i][l] == x) l -- ;
            while (r < m && g[i][r] == x) r ++ ;
            while (u >= 0 && g[u][j] == x) u -- ;
            while (d < n && g[d][j] == x) d ++ ;
            st[i][j] = r - l - 1 >= 3 || d - u - 1 >= 3;
        }

    for (int i = 0; i < n; i ++ )
    {
        for (int j = 0; j < m; j ++ )
            if (st[i][j]) cout << 0 << ' ';
            else cout << g[i][j] << ' ';
        cout << endl;
    }

    return 0;
}

  

 

  
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总结

水题，不解释

文章来源: chen-ac.blog.csdn.net，作者：辰chen，版权归原作者所有，如需转载，请联系作者。

原文链接：chen-ac.blog.csdn.net/article/details/120449665