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++ (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
static const int Max = 10000;
static const int Infty = (1<<20);
static const int W = 0;
static const int G = 1;
static const int B = 2;

int n;
vector<pair<int, int> > adj[Max];

void dijkstra(){
  priority_queue<pair<int, int> > PQ;
  int i, j, u, v, color[Max], d[Max];

  for(i = 0 ; i < n ; i++){
    d[i] = Infty;
    color[i] = W;
  }

  d[0] = 0;
  PQ.push(make_pair(0, 0));
  color[0] = G;

  while(!PQ.empty()){
    pair<int, int> f = PQ.top();
    PQ.pop();
    u = f.second;
    color[u] = B;

    if(d[u] < f.first * (-1)) continue;

    for(j = 0 ; j < adj[u].size() ; j++){
      v = adj[u][j].first;
      if(color[v] == B) continue;
      if(d[v] > d[u] + adj[u][j].second){
        d[v] = d[u] + adj[u][j].second;
        PQ.push(make_pair(d[v] * (-1), v));
        color[v] = G;
      }
    }
  }

  for(i = 0 ; i < n ; i++){
    cout << i << " " << (d[i] == Infty ? -1 : d[i]) << endl;
  }
}



int main(){
  int i, j, k, u, v, c;

  cin >> n;
  for(i = 0 ; i < n ; i++){
    cin >> u >> k;
    for(j = 0 ; j < k ; j++){
      cin >> v >> c;
      adj[u].push_back(make_pair(v, c));
    }
  }

  dijkstra();

  return 0;
}