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<stdio.h>
#include<stdlib.h>

#define N 100001
#define nil -1
#define WHITE 0
#define BLACK 1
#define INF 9999999

typedef struct{
  int prev,next,cost;
}Edge;
 
int n,d[N];
Edge e[500000];

void dijkstra(int);
void initializeSingleSource(int);

int main(){

  int i,j,u,k,c,ed=0,v,w=INF,r;
  
  scanf("%d",&n);

  for(i=0;i<n;i++){
    scanf("%d%d",&u,&k);
      for(j=0;j<k;j++){
	e[ed].prev=u;
	scanf("%d%d",&e[ed].next,&e[ed].cost);
	ed++;
      }
  }
  
  dijkstra(ed);

  for(i=0;i<n;i++) printf("%d %d\n",i,d[i]);
  
  return 0;
}


void dijkstra(int edges){

  int i,flag;

  for(i=0;i<n;i++){
   d[i]=INF;
  }
  d[0]=0;
    
  flag=1;
  while(flag!=0){
    flag=0;
    for(i=0;i<edges;i++){
      if(e[i].cost+d[e[i].prev]<d[e[i].next]){
	d[e[i].next]=e[i].cost+d[e[i].prev];
	flag=1;
      }
    }
  }
}