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>

#define INF 200000
typedef struct{
  int prev;
  int next;
  int c;
}Data;

void dijkstra(int);

int n;
Data G[500000];

int main(){
  int i,j,u,k,a,b,num = 0;

  scanf("%d",&n);
   
  for(i = 0;i < n; i++){
    scanf("%d %d",&u,&k);
    
    for(j = 0;j < k; j++){
     scanf("%d %d",&G[num].next,&G[num].c);
     G[num].prev = u;
     num++;
  }
}
  dijkstra(num);
  return 0;
}

void dijkstra(int num){
  int i,flag = 1;
  int d[10001];
  
  for(i = 0;i < n; i++){
    d[i] = INF;
  }

  d[0] = 0;
 
  while(flag == 1){
    flag = 0;

    for(i = 0; i < num; i++){
      if(d[G[i].prev] + G[i].c < d[G[i].next]){
         d[G[i].next] = G[i].c + d[G[i].prev];
	 flag = 1;
      }
    }
  }
  
  for(i = 0; i < n; i++){
    printf("%d %d\n",i,d[i]);
  
  }
}