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 MAX 100001
#define INFTY 10000000

typedef struct{
  int prev;
  int next;
  int cost;
} Edge;

Edge G[500000];
int d[MAX];

void Dijkstra(int,int);

int main(void){
  int i, j, k, n, v, u, c;
  int count=0;

  scanf("%d", &n);
  //G = (int*)malloc(sizeof(int)*n);

  /*for(i=0; i<n; i++){
    for(j=0; j<n; j++){
      G[i][j] = INFTY;
    }
    }*/

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

  Dijkstra(n,count);

  //free(G);

  return 0;
}

void Dijkstra(int n, int count){
  int i, min = 1;
  int d[MAX];

  d[0] = 0;
  for(i=1; i<n; i++){
    d[i] = INFTY;
  }

  while(min != 0){
    min = 0;

    /*for(i=0; i<n; i++){
      if(c[i] != 1 && d[i] < min){
        j = i;
        min = d[i];
      }
      }*/

    //if(min == INFTY) break;

    for(i=0; i<count; i++){
      if(d[G[i].prev] + G[i].cost < d[G[i].next]){
        d[G[i].next] = d[G[i].prev] + G[i].cost;
        min = 1;
      }
    }
  }

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

  return;
}