CCF elimination games (Full Score code + problem solving ideas + skill summary) February 2, 2015

Skill summary

Sometimes thinking about a problem , want Think more from a computer perspective
Perhaps the point by point judgment process looks very ” stupid “, But normal thinking ” First judge whether the whole string should be eliminated , Then eliminate “ The code implementation of is more cumbersome

Title Description

Their thinking

• The data range of this question is not large , You can use pure simulated violence to solve problems
• To judge whether this series should be eliminated, we can think about whether the current point should be eliminated
• Move the pointer up, down, left and right from this point , When the values are equal , The pointer moves , Judge whether three or more are equal by pointer spacing

Code implementation

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 35;

int n, m;
int a[N][N];
bool b[N][N];

int main()
{
    
    cin >> n >> m;

    for (int i = 0; i < n; i ++)
    {
    
        for (int j = 0; j < m; j ++)
        {
    
            cin >> a[i][j];
        }
    }
    
    // Traverse every point , Determine whether it will be eliminated 
    for (int i = 0; i < n; i ++) 
    {
    
        for (int j = 0; j < m; j ++)
        {
    
            int z = j, y = j, s = i, x = i; // Set up four pointers 
            while (z >= 0 && a[i][z] == a[i][j]) z --; // At the end of the cycle , The pointer points to the first position that does not meet the condition 
            while (y < m && a[i][y] == a[i][j]) y ++;
            while (s >= 0 && a[s][j] == a[i][j]) s --;
            while (x < n && a[x][j] == a[i][j]) x ++;
            if (y - z >= 4 || x - s >= 4) b[i][j] = true; // for example 3,4,5 These three spaces are exactly the same color , Then the current pointer z = 2, y = 6, So it should be >= 4
            // When there are three or more consecutive identical ones on the left, right, up and down , Then the point will be eliminated 
        }
    }
    
    for (int i = 0; i < n; i ++)
    {
    
        for (int j = 0; j < m; j ++)
        {
    
            if (!b[i][j]) cout << a[i][j] << " ";
            else cout << 0 << " ";
        }
        cout << endl;
    }
    return 0;
}