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<cstdio>
#define INF 1<<30
#define FinishFunction return 0
#define MAX 1000000
using namespace std;
    
struct edge {
  int from,to,cost,k;
};
  
edge es[MAX];
  
int main(){
  int n,u,k,E,c,v;
  E = 0;
  scanf("%d",&n);
  
  for(int i=0;i<n;i++){
    scanf("%d %d",&u,&k);
     
    for(int j=0;j<k;j++){
      scanf("%d %d",&v,&c);
      es[E].from = u;
      es[E].to = v;
      es[E++].cost = c;
    }
  }
  
  // Ika berumanfo-do   
  int d[MAX];
  for(int i=0;i<n;i++)d[i] = INF;
  d[0] = 0;
  while(true){
    bool update = false;
    for(int i=0;i<E;i++){
      edge e = es[i];
      if(d[e.from] != INF && d[e.to] > d[e.from] + e.cost){
    d[e.to] = d[e.from] + e.cost;
    update = true;
      }
    }
    if(!update)break;
  }
  for(int i=0;i<n;i++)cout << i << " " << d[i] << endl;
  FinishFunction;// return 0
}