In this HackerEarth Shortest Path Revisited problem solution In the country of HackerEarth, there are N cities and M bi - directional roads. We need to transport essential items from City 1  to all other cities. (There exists a path always)

But every road has some positive amount of Toll Charge associated with it say C (it is not same for all roads). We need to find the minimum amount of charge that it required to deliver essential items for each city.

Fortunately, to our rescue we have K special offers, which means while travelling from City 1 to any other city we can select at-most K roads and we will not be charged for using those roads. 

Can you now find the minimum charge that we have to pay to deliver essential items for each city.

(Remember we require to provide answers for each destination city separately i.e. we have K offers for every city and not as a whole)

HackerEarth Shortest Path Revisited problem solution

HackerEarth Shortest Path Revisited problem solution.

#include <bits/stdc++.h>
using namespace std;

typedef int ll;
#define int long long

const int maxn=5e5+5;
vector<pair<int,int> >graph[maxn];

int cost[maxn][20];
bool vis[maxn][20];
int n,m,k;

void djkstra() {
priority_queue<pair<int, pair<int, int> > >pq;
pq.push(make_pair(0, make_pair(1, 0)));
while (!pq.empty()) {

pair<int, pair<int, int> >;
int curr_n=curr.second.first;
int curr_s=curr.second.second;
int cst=cost[curr_n][curr_s];

for (int i=0; i<graph[curr_n].size(); i++) {
if(graph[curr_n][i].first==curr_n) continue;
int c_cst=cst+graph[curr_n][i].second;
if(cost[graph[curr_n][i].first][curr_s]==-1) {
pq.push(make_pair(c_cst*(-1), make_pair(graph[curr_n][i].first, curr_s)));
else if(cost[graph[curr_n][i].first][curr_s]>c_cst) {
pq.push(make_pair(c_cst*(-1), make_pair(graph[curr_n][i].first, curr_s)));
if(curr_s==k) continue;
if(cost[graph[curr_n][i].first][curr_s+1]==-1) {
pq.push(make_pair(c_cst*(-1), make_pair(graph[curr_n][i].first, curr_s+1)));
else if(cost[graph[curr_n][i].first][curr_s+1]>c_cst) {
pq.push(make_pair(c_cst*(-1), make_pair(graph[curr_n][i].first, curr_s+1)));


for (int i=1; i<=n; i++) {
long long ret=2e18;
for (int j=0; j<=k ; j++) {
if(cost[i][j]==-1) continue;
cout<<ret<<" ";

ll main() {

for (int i=0; i<m; i++) {
int a,b,c;


Second solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1e5 + 14, maxk = 20;

int n, m, k;
ll d[maxn][maxk];
struct E{
int u, w;
vector<E> g[maxn];
int main(){
ios::sync_with_stdio(0), cin.tie(0);
cin >> n >> m >> k;
for(int i = 0; i < m; i++){
int v, u, w;
cin >> v >> u >> w;
v--, u--;
g[v].push_back({u, w});
g[u].push_back({v, w});
memset(d, 63, sizeof d);
d[0][0] = 0;
set<pair<ll, pair<int, int> > > q({{0, {0, 0}}});
auto [v, used] = q.begin() -> second;
auto upd = [&q](int u, int used, ll nw){
if(used <= k && nw < d[u][used]){
q.erase({d[u][used], {u, used}});
d[u][used] = nw;
q.insert({d[u][used], {u, used}});
for(auto [u, w] : g[v]){
upd(u, used, d[v][used] + w);
upd(u, used + 1, d[v][used]);
for(int i = 0; i < n; i++)
cout << *min_element(d[i], d[i + 1]) << ' ';
cout << '\n';