[英]Print the values of the shortest path from any element to a given element in the matrix
[英]Matrix Traversal to print all and shortest path - infinite loop
#include<iostream>
#include<random>
#include<ctime>
#include<cstdlib>
#include<vector>
#include<algorithm>
using namespace std;
int path_checker(vector<pair<int,int> > path, int i, int j)
{
//cout<<"path_checker"<<endl;
std::vector<pair<int, int> >::iterator it;
for(it = path.begin(); it!=path.end();it++)
if(it->first == i && it->second ==j)
return 1;
return 0;
}
int isVertex(int i, int j, int n)
{
//cout<<"isVertex"<<endl;
if((i>=0) && (j>=0))
{
if((i <= n) && (j <= n))
return 1;
}
return 0;
}
void printAllPathsU(int *array, int i, int j, int n, vector<pair<int,int> > path,int index)
{
// cout<<"PrintAllPathsU_first"<<endl;
// vector<pair<int,int>> path2 = {0};
if((i == n) && (j == n))
{
if((path_checker(path,i,j)))
{
cout<<"Inside printing path"<<endl;
//vector<pair<int,int> >::iterator it;
for(int i = 0; i < path.size();i++)
cout<<"a["<<path[i].first<<"]["<<path[i].second<<"] ";
return;
}
else
{
path.push_back(make_pair(i,j));
cout<<"Inside printing path"<<endl;
//vector<pair<int,int> >::iterator it;
for(int i = 0; i < path.size();i++)
cout<<"a["<<path[i].first<<"]["<<path[i].second<<"] ";
return;
}
}
if((*((array+i*n)+j) == 1) && (!path_checker(path,i,j)))
{
//cout<<path.size()<<endl;
path.push_back(make_pair(i,j));
index++;
//len++;
if(isVertex(i,j-1,n))
printAllPathsU((int *)array,i,j-1,n,path,index);
if(isVertex(i-1,j,n))
printAllPathsU((int *)array,i-1,j,n,path,index);
if(isVertex(i,j+1,n))
printAllPathsU((int *)array,i,j+1,n,path,index);
if(isVertex(i+1,j,n))
printAllPathsU((int *)array,i+1,j,n,path,index);
}
else if((*((array+i*n)+j) == 1) && (path_checker(path,i,j)))
{
//cout<<"inside second else"<<endl;
return;
}
else if(*((array+i*n)+j) == 0)
{
// cout<<"inside third else"<<endl;
return;
}
}
void printAllPaths(int *array, int n)
{
vector<pair<int,int> > path;
//cout<<"PrintALLPaths"<<endl;
printAllPathsU(array, 0, 0, n, path, 0);
}
int main()
{ //populating matrix
int n;
cout << "Enter value of n (for n x n matrix): ";
cin >> n;
int i;
int j;
int k;
int test=1;
int array[n][n];
int randomval;
int total_elements = n*n;
int counter_0 = 0;
int max_0 = 0.2 * total_elements;
cout << "Number of zeros(20% of n*n) in matrix: ";
cout<<max_0<<endl;
int count=0;
srand(time(0));
for(i = 0;i < n;i++)
{
for(j = 0;j<n;j++)
{
array[i][j]=-1;
}
}
while(count < total_elements)
{
i=rand()%n;
j=rand()%n;
if(array[i][j]==1 || array[i][j]==0)
{
continue;
}
else if(array[i][j] == -1)
{
count+=1;
if(i==0 && j==0)
{
array[i][j]=1;
}
else if (counter_0 < max_0)
{
counter_0+=1;
array[i][j] = 0;
}
else if(counter_0 >= max_0)
{
test+=1;
array[i][j] = 1;
}
}
else{continue;}
}
cout<<"# of 1s:"<<test<<" & # of 0s:"<<counter_0<<endl;
cout<<"Elements Populated:"<<count<<endl;
cout<<"Total Elements in matrix:"<<total_elements<<endl;
if(counter_0 < max_0)
{ cout<<"adding more zeros"<<endl;
while(k < (max_0 - counter_0))
{
i = rand()%n;
j = rand()%n;
if(array[i][j] == 0)
{
}
else
{
array[i][j] = 0;
k+=1;
}
}
}
for(i = 0;i < n;i++)
{
for(j = 0;j<n;j++)
{
cout<<array[i][j]<<" ";
}
cout<<endl;
}
//printing paths
if(array[1][0]==0 && array[0][1]==0)
{
cout<<"No Possible paths homie #snorlaxiseverywhere";
}else
{
//printing paths
printAllPaths((int *)array, n);
}
return 0;
}
问题是遍历填充了1和0的* n矩阵,矩阵中的0的数量是元素总数的20%,并打印所有路径,然后打印来自[0] [0]的最短路径]到[n] [n],使用四个方向:上,下,左,右。 到目前为止,我已尝试实现打印所有可能的路径。 但是,我陷入了无限循环中
else if((*((array+i*n)+j) == 1) && (!path_checker(path,i,j)))
{
...
}
我讨论path.size()来检查每个实例中的大小变化,输出有点像:
35
48
37
...and so on infinitely (values ranging approx between 30-50)
任何想法如何纠正这个?
编辑:更改了一些逻辑,而不是无限循环。 但所有函数调用都通过函数内的“second else”和“third else”退出 - printAllPathsU(...)。
谢谢!
只要问题是如何“打印所有路径”,你应该有一个无限循环,因为有无限多的路径。 如果问题是如何打印所有非自相交路径,那么这个改述就会提示如何解决问题。
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