Problem:
Single Source Shortest Path

For a given weighted graph G(V, E) and a source r, find the source shortest path to each vertex from the source (SSSP: Single Source Shortest Path).

Input

An edge-weighted graph G (V, E) and the source r.

|V| |E| r
s0 t0 d0
s1 t1 d1
:
s|E|-1 t|E|-1 d|E|-1

|V| is the number of vertices and |E| is the number of edges in G. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively. r is the source of the graph.

si and ti represent source and target vertices of i-th edge (directed) and di represents the cost of the i-th edge.

Output

Print the costs of SSSP in the following format.

c0
c1
:
c|V|-1

The output consists of |V| lines. Print the cost of the shortest path from the source r to each vertex 0, 1, ... |V|-1 in order. If there is no path from the source to a vertex, print INF.

Constraints

1 ≤ |V| ≤ 100000

0 ≤ di ≤ 10000

0 ≤ |E| ≤ 500000

There are no parallel edges

There are no self-loops

Sample Input 1

4 5 0
0 1 1
0 2 4
1 2 2
2 3 1
1 3 5

Sample Output 1

0
1
3
4

Sample Input 2

4 6 1
0 1 1
0 2 4
2 0 1
1 2 2
3 1 1
3 2 5

Sample Output 2

3
0
2
INF

Solution in C++11 (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, a, n) for(int i = a; i < n; i++)
#define repb(i, a, b) for(int i = a; i >= b; i--)
#define all(a) a.begin(), a.end()
#define o(a) cout << a << endl
#define int long long
#define fi first
#define se second
using namespace std;
typedef pair<int, int> P;

struct edge{
    int to, cost;
    edge(int to, int cost):to(to), cost(cost){}
};

const int INF = 1e9;
int v, e, r;
int d[100010];
vector<edge> G[100010];

void dijkstra(int s){
    priority_queue<P, vector<P>, greater<P> > q;
    fill(d, d + v, INF);
    q.push(P(0, s));
    d[s] = 0;
    while(q.size()){
        P p = q.top(); q.pop();
        int now = p.se;
        int nowCost = p.fi;
        if(nowCost > d[now]) continue;
        for(auto i: G[now]){
            int next = i.to;
            int cost = i.cost;
            if(d[next] > nowCost + cost){
                d[next] = nowCost + cost;
                q.push(P(d[next], next));
            }
        }
    }
}

signed main(){
    cin >> v >> e >> r;
    rep(i, 0, e){
        int s, t, d;
        cin >> s >> t >> d;
        G[s]. push_back(edge(t, d));
        // G[t]. push_back(edge(s, d));
    }
    dijkstra(r);
    rep(i, 0, v){
        if(d[i] == INF) o("INF");
        else o(d[i]);
    }
}