C++ Program to Determine Round Trip Possibility from a City

Suppose, there are n cities and m roads are connecting them. Each road is unidirectional, and it takes a particular amount of time to reach from the source city to the destination city. The information of the roads is given in the array roads where each element is of the format (source, destination, time). Now, a person is traveling from one city to another city and the trip has to be a round-trip. A trip can be called round-trip when the person starts from a particular city, goes through one or more roads, and finishes the trip in the same city. So for each city, we have to determine if a round-trip is possible from that particular city. If it is possible, print the time required to perform the round-trip or else print -1.

So, if the input is like n = 4, m = 4, roads = {{1, 2, 5}, {2, 3, 8}, {3, 4, 7}, {4, 1, 6}}, then the output will be: 26 26 26 26. From each city, it takes time 26 to perform a round-trip.

To solve this, we will follow these steps −

Define one 2D array graph(n) of pairs
for initialize i := 0, when i < m, update (increase i by 1), do:
   x := first value of roads[i]
   y := second value of roads[i]
   z := third value of roads[i]
   decrease x and y by 1
   insert pair (y, z) at the end of graph[x]
for initialize i := 0, when i < n, update (increase i by 1), do:
   q := a new priority queue
   Define an array dst
   insert pair (0, i) at the top of q
   while size of q is non-zero, do:
      pair p := top value of q
      delete the top element from q
      dt := first value of p
      curr := second value of p
      if dst[curr] is same as 0, then:
         dst[curr] := dt
         Come out from the loop
      if dst[curr] is not equal to -1, then:
         Ignore following part, skip to the next iteration
      dst[curr] := dt
      for element next in graph[curr], do:
         tp := first value of next
         cst := second value of next
         insert pair(dt + cst, tp) at the top of q
   if dst[i] is same as 0, then:
      dst[i] := -1
   print(dst[i])

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int modval = (int) 1e9 + 7;
#define N 100
void solve(int n, int m, vector<tuple<int, int, int>> roads ) {
   vector<vector<pair<int, int>>> graph(n);
   for(int i = 0; i < m; i++) {
      int x, y, z;
      tie(x, y, z) = roads[i];
      x--; y--;
      graph[x].emplace_back(y, z);
   }
   for(int i = 0; i < n; i++) {
      priority_queue<pair<int, int>> q;
      vector<int> dst(n, -1);
      q.emplace(0, i);
      while(q.size()){
         pair<int, int> p = q.top();
         q.pop();
         int curr, dt;
         tie(dt, curr) = p;
         if(dst[curr] == 0) {
            dst[curr] = dt;
            break;
         }
         if(dst[curr] != -1)
            continue;
         dst[curr] = dt;
         for(auto next : graph[curr]){
            int tp, cst;
            tie(tp, cst) = next;
            q.emplace(dt + cst, tp);
         }
      }
      if(dst[i] == 0)
      dst[i] = -1;
      cout<< dst[i]<< endl;
   }
}
int main() {
   int n = 4, m = 4;
   vector<tuple<int, int, int>> roads = {{1, 2, 5}, {2, 3, 8}, {3, 4, 7}, {4, 1, 6}};
   solve(n, m, roads);
   return 0;
}

Input

4, 4, {{1, 2, 5}, {2, 3, 8}, {3, 4, 7}, {4, 1, 6}}

Output

26
26
26
26