Есть текст программы, и нужно переделать чтоб исходные данные брались не с файла, а пользователь сам вводил с клавиатуры!!
C++:
#include <iostream.h>
#include <fstream.h>
#include<conio.h>
main(int argc, char *argv[]) {
//definition of the variables.
int mtx1[20][20], mtx2[20][20], mtx3[20][20], mtx4[20][20];
int n, i, j, k;
int nlink=2, max=9999;
//exit if the number of arguments is not 1.
if(argc != 2) {
cerr << "Usage: shimbel <filename>\n";
return 1;
}
//open the input file. exit if an error occurs.
ifstream fin(argv[1]);
if(!fin) {
cerr << "Can't open file!\n";
return 1;
}
//read the input file.
fin >> n; //read the number of rows and columns.
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
fin >> mtx1[i][j]; //read the elements of the matrix.
}
}
fin.close(); //close the file.
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
mtx2[i][j] = mtx1[i][j]; //copy elements of mtx1 into mtx2.
}
}
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
mtx4[i][j] = 9999; //set elements of mtx4 as 9999.
}
}
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
if(i == j) {
mtx4[i][j] = 0; //put 0 in the diagonal elements.
} else if(mtx1[i][j] == 1) {
mtx4[i][j] = 1; //put 1 if a linkage exists between the nodal pair.
}
}
}
nlink = 2; //start with 2 step path.
while(1) { //endless loop
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
mtx3[i][j] = 0; //set elements of mtx3 as 0.
}
}
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
for(k=1; k<=n; k++) {
mtx3[i][j] += mtx2[i][k] * mtx1[k][j]; //multiply the matrices.
}
if (mtx3[i][j] != 0){
if(mtx4[i][j] > nlink) {
mtx4[i][j] = nlink; //put the step number in the nodal pair when it becomes non-zero firstly.
}
}
}
}
//get the maximum element of mtx4.
max = 0;
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
if(mtx4[i][j] > max) {
max = mtx4[i][j];
}
}
}
if(max < 9999) break; //if the maximun is not 9999, exit the loop.
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
mtx2[i][j] = mtx3[i][j]; //copy elements of mtx3 into mtx2.
}
}
nlink++; //increment nlink.
} //end of the loop.
//output the results.
cout << n << "\n"; //number of rows and columuns
for(j=1; j<=n; j++) {
for(i=1; i<=n; i++) {
cout << mtx4[i][j]; //elements of the matrix
if(i < n) {
cout << "\t";
}
}
cout << "\n";
}
return 0;
} //end of the program.