Problem:
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Single Source Shortest Path II

For a given weighted graph $G = (V, E)$, find the shortest path from a source to each vertex. For each vertex $u$, print the total weight of edges on the shortest path from vertex $0$ to $u$.

Input

In the first line, an integer $n$ denoting the number of vertices in $G$ is given. In the following $n$ lines, adjacency lists for each vertex $u$ are respectively given in the following format:

$u$ $k$ $v_1$ $c_1$ $v_2$ $c_2$ ... $v_k$ $c_k$

Vertices in $G$ are named with IDs $0, 1, ..., n-1$. $u$ is ID of the target vertex and $k$ denotes its degree. $v_i (i = 1, 2, ... k)$ denote IDs of vertices adjacent to $u$ and $c_i$ denotes the weight of a directed edge connecting $u$ and $v_i$ (from $u$ to $v_i$).

Output

For each vertex, print its ID and the distance separated by a space character in a line respectively. Print in order of vertex IDs.

Constraints

$1 \leq n \leq 10,000$

$0 \leq c_i \leq 100,000$

$|E| < 500,000$

All vertices are reachable from vertex $0$

Sample Input 1

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

Sample Output 1

0 0
1 2
2 2
3 1
4 3

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in C++14 (judge verdict: Accepted)

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

vector<int> d;
vector<vector<pair<int, int>>> edge;

void dijkstra(int s){
	priority_queue<pair<int, int>> pq;
	pq.push(make_pair(0, s));

	while(!pq.empty()){
		int cost = pq.top().first;
		int v = pq.top().second;
		pq.pop();

		if(d[v] != 1) continue;

		d[v] = cost;

		for(auto e : edge[v]){
			if(d[e.second] == 1){
				pq.push(make_pair(cost + e.first, e.second));
			}
		}
	}
}

int main(){
	int n; cin >> n;

	d.resize(n, 1);
	edge.resize(n);
	for(int i=0; i<n; ++i){
		int v, k; cin >> v >> k;
		for(int l=0; l<k; ++l){
			int a, c; cin >> a >> c;
			edge[v].push_back(make_pair(-c, a));
		}
	}

	dijkstra(0);

	for(int i=0; i<n; ++i){
		cout << i << " " << -d[i] << endl;
	}
}