找到列中的最大元素，同时也是矩阵中行中的最小元素

Question

Write a program to enter integers in a matrix of size L*C (L: row and C: column) then determine and display the value as well as the position (i,j) of all the elements which are both a minimum on their line and a maximum on their column. If no Min-Max does not exist, we display the following message "The matrix does not contain any Min-Max.". Example 1: 1 2 3 4 5 6 7 8 9 13 15 16 Tab[4,1]=13 Example 2: 17 11 3 4 5 6 7 8 9 15 10 16 The matrix contains no Min-Max. i tried this code and it didnt work

 int L , C , i , j ,maxc,minl ;
    int Tab[20][20];
printf("Introduire le nombre des lignes du matrice (MAX 20): ");
        scanf("%d",&L);
        printf("Introduire le nombre des colonnes du matrice (MAX 20): ");
        scanf("%d",&C);
        for(i=0; i<L; i++)
        {
            for(j=0; j<C; j++)
            {
            printf("Donner l'element (%d,%d): ",i+1,j+1);
            scanf("%d", &Tab[i][j]);
            }
        }


        for(j=0 ;j<C;j++)
        {
            maxc=Tab[0][j];
            for(i=0;i<L;i++)
            {
                minl=Tab[i][0];
               if(Tab[i][j]>maxc && Tab[i][j]<minl )
               {
                 printf("Tab[%d,%d]=%d ",i+1,j+1,Tab[i][j]);
               }
               else
               {
                   printf("La matrice ne contient aucun Min-Max.");
               }
            }
        }

    return 0;
}

Answer 1

I'm not sure why are you creating a matrix of size 20x20 when you only need L rows and C columns? You can just declare it as

int Tab[L][C];

just after you have the value of L and C variables returned from scanf .

As for the algorithm, it does not work because when using maxc=Tab[0][j] , you are fixing the row index to zero (first row): for every column you are updating maxc to the corresponding row items value stored in the first row. The same for minl=Tab[i][0] : for every column, and for every row, you update maxc to the items of row index i which are stored in the first column. In other words maxc can assume only values of the first row, and minl can assume only values of the first column.

Then, when doing if(Tab[i][j]>maxc && Tab[i][j]<minl) , maxc and minl can't match: as for example at the first iteration ( j=0 ) maxc is stuck to the item [0,0], while minl goes in the positions [0,0], [1,0], [2,0], [3,0]; then, when j=0 and i=3 (last step of first iteration of the inner cycle), you have maxc=[0,0] , minl=[3,0] , and Tab[i][j]=[3][0] .

Instead you have to compare the maximum and minimum values against the other values in the rows and columns, before updating maxc and minl : otherwise, they could just randomly match, depending on the matrix structure itself, and it won't even match when they contains the real minimum and maximum values. Also, you should reset minl and maxc every time you exit from the inner cycle, in order to avoid dealing with minimum and maximum values of the previous row.

As for the following code it should work as expected, I just did the inverse, by searching the minimum value for every row and saving its index in the a variable (because you will lose the index if the minimum value is not the latest one); then, when you have the minimum value for a specific row, you iterate that column with the index a and search if it exist an element which is bigger: in case it exists, it updates maxc .

I would also suggest to avoid inverting i and j indexes in order to avoid confusion (in your code you was using j for columns and i for rows, but during the insertion of the values in the matrix you used i for columns and j for rows).

int L, C, min, max;

    printf("Introduire le nombre des lignes du matrice:\n");
    scanf("%d", &L);
    printf("Introduire le nombre des colonnes du matrice:\n");
    scanf("%d", &C);
        
    int Tab[L][C]; //Declare it here with the right size    
    int i; //Index for iterating over rows
    int j; //Index for iterating over columns
    int a; //Index of the column with the minimum value (for each row)
    int k; //Index for iterating the column that contains the min row value.
    bool found = false; 
    
    for(i=0; i<L; i++){
        for(j=0; j<C; j++){
            printf("Donner l'element (%d, %d): ", i, j);
            scanf("%d", &Tab[i][j]);
            
            if(j==0){
                min = max = Tab[i][j];
            } else {
                //Store the minimum and max values while user input them
                if(Tab[i][j]<min) min = Tab[i][j]; 
                if(Tab[i][j]>max) max = Tab[i][j];
            }
        }
    }
    
    for(i=0; i<L; i++){
        //You need to reset 'maxc' and 'minl' for every row, otherwise you will compare it with the minimum and maximum values of the previous rows
        int maxc = min-1;
        int minl = max+1;
        for(j=0; j<C; j++){ 
            if(Tab[i][j]<minl){
                minl = Tab[i][j];
                a = j;
            }
            if(j==C-1){ //Here 'a' contains the row index of the minimum element of the row 'i'
                for(k=0; k<L; k++){ //Iterating the column with 'a' index: we search for the maximum value in that column
                    if(Tab[k][a]>maxc) maxc = Tab[k][a];
                    if(k==L-1){ // Here 'a' contains the row index of the minimum element of the row 'i' (minl variable), and also here the maxc variable contains the maximum element of the row 'i'.
                        if(minl == maxc){
                            printf("Tab[%d,%d]=%d ", i, a, Tab[i][a]);
                            found = true;
                        }
                    }
                }
            }
        }
    }
    if(!found) printf("La matrice ne contient aucun Min-Max.");