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Using the Breadth-First Search Algorithm, determine the minimum number of edges that it would require to reach

jekas [21]

Answer:

The algorithm is given below.

#include <iostream>

#include <vector>

#include <utility>

#include <algorithm>

using namespace std;

const int MAX = 1e4 + 5;

int id[MAX], nodes, edges;

pair <long long, pair<int, int> > p[MAX];

void initialize()

{

for(int i = 0;i < MAX;++i)

id[i] = i;

}

int root(int x)

{

while(id[x] != x)

{

id[x] = id[id[x]];

x = id[x];

}

return x;

}

void union1(int x, int y)

{

int p = root(x);

int q = root(y);

id[p] = id[q];

}

long long kruskal(pair<long long, pair<int, int> > p[])

{

int x, y;

long long cost, minimumCost = 0;

for(int i = 0;i < edges;++i)

{

// Selecting edges one by one in increasing order from the beginning

x = p[i].second.first;

y = p[i].second.second;

cost = p[i].first;

// Check if the selected edge is creating a cycle or not

if(root(x) != root(y))

{

minimumCost += cost;

union1(x, y);

}

return minimumCost;

}

int main()

{

int x, y;

long long weight, cost, minimumCost;

initialize();

cin >> nodes >> edges;

for(int i = 0;i < edges;++i)

{

cin >> x >> y >> weight;

p[i] = make_pair(weight, make_pair(x, y));

}

// Sort the edges in the ascending order

sort(p, p + edges);

minimumCost = kruskal(p);

cout << minimumCost << endl;

return 0;

}

8 0

3 years ago

The manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .

Misha Larkins [42]

Answer:

0.1507 or 15.07%.

Step-by-step explanation:

We have been given that the manufacturing of a ball bearing is normally distributed with a mean diameter of 22 millimeters and a standard deviation of .016 millimeters. To be acceptable the diameter needs to be between 21.97 and 22.03 millimeters.

First of all, we will find z-scores for data points using z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,

z = z-score,

x = Sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

$z=\frac{21.97-22}{0.016}$

$z=\frac{-0.03}{0.016}$

$z=-0.1875$

Let us find z-score of data point 22.03.

$z=\frac{22.03-22}{0.016}$

$z=\frac{0.03}{0.016}$

$z=0.1875$

Using probability formula $P(a$ , we will get:

$P(-0.1875$

Therefore, the probability that a randomly selected ball bearing will be acceptable is 0.1507 or 15.07%.

6 0

3 years ago

Y

qaws [65]

Answer:

(-2, 0.5) first

Step-by-step explanation:

X (-4, 1) → X' (-4*0.5, 1*0.5)

X'(-2, .5)

5 0

2 years ago

The 70th term of the arithmetic sequence the 70th term of the arithmetic sequence 18,34,50,...

Sedbober [7]

Answer:1122

Step-by-step explanation:

Hve to work out the general equation first by calculating the common difference-16.

3 0

2 years ago

What is the answer and the work i have to do to get to it

harina [27]

Answer:

B

Step-by-step explanation:

4 0

2 years ago