Cplex c++ loop for

Cplex c++ loop for

本文关键字:for loop c++ Cplex      更新时间:2023-10-16

我需要在visual c++中使用可调用库编写一个复杂的程序。

我需要这样使用numrows和numcols。

我只把我的程序的开始,因为我的问题是在开始。我的程序崩溃了,我找到了。它发生在循环之后,以增加numrows。看起来它不能跳出循环std::cout <<NUMROWS & lt; & lt;"answers";永远不会出现。如果我在循环中写入它,我将看到值,但之后看不到。我找不到原因。你知道为什么吗?

感谢
//subfunction pk, xik, yjk
int P(int k){
return k-1;
}
int X(int i, int k){
int p, n;
return p + (i-1)*(n-1) + (k-1);
}
int Y(int j, int k){
int p, n, m;
return p + n*p + (j-1)*(m-1) + (k-1);
}
int main (int argc, char **argv){
srand(time(0));
int status = 0;
int project = 4; 
int employee = 5; 
int time = 5; 
int empl [] = {2, 2, 2, 3};
int meet [] = {2, 4, 3, 3};
int n, m, k, p;
n=5;
m=5;
p=4;
int NUMCOLS=n+m;
CPXENVptr env = NULL;
CPXLPptr  lp  = NULL;
double *obj = new double [NUMCOLS]; 
//Objective function
int profit [] = {10, 20, 5, 15};
for (int i=0; i<project; i++){
        obj[i]=profit[i];
}
int      solstat;
double   objval;
double *lb = new double [NUMCOLS];
double *ub = new double [NUMCOLS];
double *x  = new double [NUMCOLS];
int    *matbeg = new int [NUMCOLS];
int    *matcnt = new int [NUMCOLS];
char   *ctype = new char [NUMCOLS];
int **F = new int*[employee]; 
for(int a = 0; a < employee; a++){      
    F[a] = new int [time];
    for(int b = 0; b < time; b++){
        F[a][b]=rand()%2;
        std::cout << F[a][b] << ", ";
    }
}
int NUMROWS=0;  
for(int i=1; i<=n; i++){ //Each xi      
    for(int j=1; j<=m; j++){ //Each yj
        if(F[i][j] ==0){
            NUMROWS++;
        } 
    }
}std::cout << NUMROWS << "and ";
double *rhs = new double [NUMROWS+1];
char   *sense = new char [NUMROWS+1];
for(int i=0; i < NUMROWS; i++) { //Each row
    rhs[i]=1;
    sense[i]='L';
}
int num_entries=-1;
for(int i=0; i < n; i++) {
    for(int j=0; j < m; j++) {
        if(F[i+1][j+1] ==0) { //1st constraint (pk, xik, yjk)
            num_entries++;
            num_entries++;
            num_entries++;
            std::cout << "try ";
        }
    }
}
int NUMNZ = NUMROWS*NUMCOLS;
int    *matind = new int [2*NUMROWS+m];
double *matval  = new double [2*NUMROWS+m];
matrix_entry *M = new matrix_entry  [num_entries+1+m];
num_entries=-1;
int row=-1;
for(int i=0; i < n; i++) {//1st constraint
    for(int j=0; j < m; j++) {
        if(F[i+1][j+1] ==0) {
            row++;
            num_entries++;
            M[num_entries].col=P(project); //pk
            M[num_entries].row=row;
            M[num_entries].val=-1;
            num_entries++;
            M[num_entries].col=X(i,project); //xik
            M[num_entries].row=row;
            M[num_entries].val=1;
            num_entries++;
            M[num_entries].col=Y(j,project); //yik
            M[num_entries].row=row;
            M[num_entries].val=1;
        }
    }
}

您有以下声明:

int employee = 5; 
int time = 5; 
n=5;
m=5;

然后使用——

初始化F数组
for(int a = 0; a < employee; a++){      
    F[a] = new int [time];
    for(int b = 0; b < time; b++){
         F[a][b]=rand()%2;

然后使用——

访问它
for(int i=1; i<=n; i++){ //Each xi      
    for(int j=1; j<=m; j++){ //Each yj
        if(F[i][j] ==0){

所以你在1的范围之外。循环应该是

for(int i=0 i<n; i++){ //Each xi      
    for(int j=0; j<m; j++){ //Each yj

建议:

  1. 请使用std::vector或其他容器(Basile Starynkevitch建议)

  2. 请给你的变量起个好名字(容易理解和记忆)

  3. 请在循环中使用的变量保持一致,您将更少忘记边界