Please explain Sending email more clearly for me please there's a picture of the decription and also please comment on each line if possible on the code below to for me to have a better understanding of what each line is doing. It is in C++. Please also give the space and time complexity and the reason why. #include #include #include #include #include using namespace std; #define pii pair #define P pair #define F first #define S second #define INF 0x3F3F3F3F3F3F3F3F const int maxn = 20000; vector G[maxn]; long long dist[maxn]; int main() { int N, n, m, S, T; int u, v, w; cin >> N; for (int Case=1; Case<=N; Case++){ for (int i=0; i> n >> m >> S >> T; for (int i=0; i> u >> v >> w; G[u].push_back({v, w}); G[v].push_back({u, w}); } // Dijkstra - priority_queue memset(dist, 0x3F, sizeof(dist)); priority_queue, greater > pq; dist[S] = 0; pq.push({0, S}); while (!pq.empty()){ P p = pq.top(); pq.pop(); if (dist[p.S] < p.F) continue; for (auto nxt: G[p.S]){ if (dist[nxt.F] > dist[p.S] + nxt.S){ dist[nxt.F] = dist[p.S] + nxt.S; pq.push({dist[nxt.F], nxt.F}); } } } cout << "Case #" << Case << ": "; if (dist[T] == INF) cout << "unreachable\n"; else cout << dist[T] << '\n'; } return 0; }
Please explain Sending email more clearly for me please there's a picture of the decription and also please comment on each line if possible on the code below to for me to have a better understanding of what each line is doing. It is in C++. Please also give the space and time complexity and the reason why.
#include <iostream>
#include <cstring>
#include <
#include <queue>
#include <utility>
using namespace std;
#define pii pair<int, int>
#define P pair<long long, int>
#define F first
#define S second
#define INF 0x3F3F3F3F3F3F3F3F
const int maxn = 20000;
vector<pii> G[maxn];
long long dist[maxn];
int main() {
int N, n, m, S, T;
int u, v, w;
cin >> N;
for (int Case=1; Case<=N; Case++){
for (int i=0; i<maxn; i++) G[i].clear();
cin >> n >> m >> S >> T;
for (int i=0; i<m; i++){
cin >> u >> v >> w;
G[u].push_back({v, w});
G[v].push_back({u, w});
}
// Dijkstra - priority_queue
memset(dist, 0x3F, sizeof(dist));
priority_queue<P, vector<P>, greater<P> > pq;
dist[S] = 0;
pq.push({0, S});
while (!pq.empty()){
P p = pq.top();
pq.pop();
if (dist[p.S] < p.F) continue;
for (auto nxt: G[p.S]){
if (dist[nxt.F] > dist[p.S] + nxt.S){
dist[nxt.F] = dist[p.S] + nxt.S;
pq.push({dist[nxt.F], nxt.F});
}
}
}
cout << "Case #" << Case << ": ";
if (dist[T] == INF) cout << "unreachable\n";
else cout << dist[T] << '\n';
}
return 0;
}
![Sending email
Description
"A new internet watchdog is creating a stir in
Springfield. Mr. X, if that is his real name, has
come up with a sensational scoop."
Kent Brockman
There are n SMTP servers connected by network cables. Each of the m cables connects two computers and has a certain latency measured in milliseconds required to send an email message. What is th
e shortest time required to send a message from server S to server T along a sequence of cables?
Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containing n(2 ≤ n ≤ 20000), m(0 ≤m ≤ 50000), S(0 ≤S<n) and T(0 ≤ T <n). $ =
T. The next m lines will each contain 3 integers: 2 different servers (in the range [0, n - 1]) that are connected by a biderectional cable and the latency, w, along with cable ($0 \le w \le 10000).
Output
For each test case, output the line 'Case #x:' followed by the number of milliseconds required to send a message from S to T. Print 'unreachable' if there is no route from S to T.
Sample Input 1 ]
Sample Output 1
3
Case #1: 100
Case #2: 150
2101
Case #3: unreachable
0 1 100
3 3 20
0 1 100
0 2 200
1 2 50
2001](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb42a9107-65e6-4614-a858-595dd78b3016%2F919c8d47-f477-46b7-998c-075eebc8c975%2Fzfpzd7_processed.png&w=3840&q=75)
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