Floyd algorithm implemented with (const vector& t : flights), what are the values stored on 't'
I came across this implementation of floyd, but i have a question as to what 't' stores as values, seeing that flights is a vector of vectors. (I understand how the algorithm works.)
//vector<vector<int>>& flights
//vector<vector<int>> vec(n, vector<int>(k + 1));
for (int i = 1; i <= k; i++)
{
for (int j = 0; j < n; j++)
{
vec[j][i] = vec[j][i - 1];
for (const vector<int>& t : flights)
{
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
}
}
c++ vector
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
asked Nov 12 '18 at 23:02
DanielDaniel
31
add a comment |
I came across this implementation of floyd, but i have a question as to what 't' stores as values, seeing that flights is a vector of vectors. (I understand how the algorithm works.)
//vector<vector<int>>& flights
//vector<vector<int>> vec(n, vector<int>(k + 1));
for (int i = 1; i <= k; i++)
{
for (int j = 0; j < n; j++)
{
vec[j][i] = vec[j][i - 1];
for (const vector<int>& t : flights)
{
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
}
}
c++ vector
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
asked Nov 12 '18 at 23:02
DanielDaniel
31
add a comment |
I came across this implementation of floyd, but i have a question as to what 't' stores as values, seeing that flights is a vector of vectors. (I understand how the algorithm works.)
//vector<vector<int>>& flights
//vector<vector<int>> vec(n, vector<int>(k + 1));
for (int i = 1; i <= k; i++)
{
for (int j = 0; j < n; j++)
{
vec[j][i] = vec[j][i - 1];
for (const vector<int>& t : flights)
{
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
}
}
c++ vector
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
asked Nov 12 '18 at 23:02
DanielDaniel
31
I came across this implementation of floyd, but i have a question as to what 't' stores as values, seeing that flights is a vector of vectors. (I understand how the algorithm works.)
//vector<vector<int>>& flights
//vector<vector<int>> vec(n, vector<int>(k + 1));
for (int i = 1; i <= k; i++)
{
for (int j = 0; j < n; j++)
{
vec[j][i] = vec[j][i - 1];
for (const vector<int>& t : flights)
{
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
}
}
c++ vector
c++ vector
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
asked Nov 12 '18 at 23:02
DanielDaniel
31
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
asked Nov 12 '18 at 23:02
DanielDaniel
31
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
edited Nov 13 '18 at 0:21
AS Mackay
1,9894819
1,9894819
asked Nov 12 '18 at 23:02
DanielDaniel
31
asked Nov 12 '18 at 23:02
DanielDaniel
31
asked Nov 12 '18 at 23:02
DanielDaniel
31
31
add a comment |
add a comment |
1 Answer
1
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oldest
votes
This syntax is a range-based loop, introduced in C++11. Here, t is just a const reference to an element of flights. The loop will visit each element of flights in order, and you can use the identifier t to reference the current element.
The loop is roughly equivalent to the following pre-C++11 syntax:
for(std::vector<std::vector<int>>::const_iterator it = flights.begin(); it != flights.end(); ++it)
{
const vector<int>& t = *it;
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This syntax is a range-based loop, introduced in C++11. Here, t is just a const reference to an element of flights. The loop will visit each element of flights in order, and you can use the identifier t to reference the current element.
The loop is roughly equivalent to the following pre-C++11 syntax:
for(std::vector<std::vector<int>>::const_iterator it = flights.begin(); it != flights.end(); ++it)
{
const vector<int>& t = *it;
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
add a comment |
This syntax is a range-based loop, introduced in C++11. Here, t is just a const reference to an element of flights. The loop will visit each element of flights in order, and you can use the identifier t to reference the current element.
The loop is roughly equivalent to the following pre-C++11 syntax:
for(std::vector<std::vector<int>>::const_iterator it = flights.begin(); it != flights.end(); ++it)
{
const vector<int>& t = *it;
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
add a comment |
This syntax is a range-based loop, introduced in C++11. Here, t is just a const reference to an element of flights. The loop will visit each element of flights in order, and you can use the identifier t to reference the current element.
The loop is roughly equivalent to the following pre-C++11 syntax:
for(std::vector<std::vector<int>>::const_iterator it = flights.begin(); it != flights.end(); ++it)
{
const vector<int>& t = *it;
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
This syntax is a range-based loop, introduced in C++11. Here, t is just a const reference to an element of flights. The loop will visit each element of flights in order, and you can use the identifier t to reference the current element.
The loop is roughly equivalent to the following pre-C++11 syntax:
for(std::vector<std::vector<int>>::const_iterator it = flights.begin(); it != flights.end(); ++it)
{
const vector<int>& t = *it;
vec[t[1]][i] = min(vec[t[1]][i], vec[t[0]][i - 1] + t[2]);
}
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
answered Nov 12 '18 at 23:11
paddypaddy
42.7k53176
42.7k53176
add a comment |
add a comment |
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